Chapter 12 — LIMITS AND DERIVATIVES

Concept Theory

A function is said to be continuous at a point if its limit at that point is equal to its actual value. Polynomial functions are continuous for all real values of \(x\). Therefore, limits of polynomials can be evaluated using direct substitution.

Solution Roadmap

• Identify the type of function (Polynomial → Continuous)
• Apply direct substitution
• Simplify the expression

Solution

Since \(x+3\) is a polynomial, it is continuous at \(x = 3\).

\[ \begin{aligned} \lim_{x\rightarrow 3}(x+3) &= 3 + 3 \ &= 6 \end{aligned} \]

Hence, the limit is 6.

Key Insight

For continuous functions, limit = function value. No special techniques are required.

Exam Significance

• Very common in CBSE board exams (1–2 marks direct question)
• Forms the base concept for JEE & NEET limits
• Helps identify when substitution is valid vs when advanced methods are needed

Concept Theory

Linear functions of the form \(ax + b\) are continuous for all real values of \(x\). Hence, their limits can always be evaluated using direct substitution.

Also note that \(\dfrac{22}{7}\) is a rational approximation of \(\pi\), but it is not exactly equal to \(\pi\). This means the result will be a small non-zero value.

Solution Roadmap

• Recognize the function is linear → continuous
• Apply direct substitution \(x = \pi\)
• Simplify the numerical expression

Solution

Since the function is continuous at \(x = \pi\), we directly substitute:

\[ \begin{aligned} \lim_{x\rightarrow \pi}\left(x-\dfrac{22}{7}\right) &= \pi - \dfrac{22}{7} \end{aligned} \]

This value is slightly positive because \(\pi \approx 3.1416\) and \(\dfrac{22}{7} \approx 3.1429\).

Key Insight

Even though \(\dfrac{22}{7}\) is close to \(\pi\), limits depend on exact values. Small numerical differences are preserved in limit evaluation.

Exam Significance

• Tests understanding of continuity + substitution principle
• Common conceptual trap in JEE/NEET where approximations are misused
• Strengthens clarity between exact vs approximate values

Concept Theory

The expression \(\pi r^2\) represents the area of a circle with radius \(r\). It is a polynomial function in \(r\) multiplied by a constant \(\pi\), hence it is continuous for all real values of \(r\).

For continuous functions, the limit at a point is equal to the function’s value at that point.

Solution Roadmap

• Identify the function as continuous
• Apply direct substitution \(r = 1\)
• Interpret result geometrically (area of unit circle)

Solution

Since \(\pi r^2\) is continuous at \(r = 1\), we directly substitute:

\[ \begin{aligned} \lim_{r\rightarrow 1} \pi r^2 &= \pi (1)^2 \ &= \pi \end{aligned} \]

Hence, the required limit is \(\pi\).

Key Insight

This limit represents the area of a unit circle. Whenever radius approaches 1, the area approaches \(\pi\).

Exam Significance

• Direct application of continuity + substitution
• Links algebra with geometric interpretation
• Frequently used concept in JEE/NEET (area-based limits)

Concept Theory

A rational function \(\dfrac{p(x)}{q(x)}\) is continuous at a point if the denominator is non-zero at that point. In such cases, limits can be evaluated using direct substitution.

Discontinuity occurs only when the denominator becomes zero.

Solution Roadmap

• Check denominator at \(x = 4\)
• If non-zero → function is continuous
• Apply direct substitution

Solution

Evaluate the denominator at \(x = 4\):

\[ x - 2 = 4 - 2 = 2 \neq 0 \]

Since the denominator is non-zero, the function is continuous at \(x = 4\). Therefore, we substitute directly:

\[ \begin{aligned} \lim_{x\rightarrow 4}\dfrac{4x+3}{x-2} &= \dfrac{4(4)+3}{4-2} \ &= \dfrac{16+3}{2} \ &= \dfrac{19}{2} \end{aligned} \]

Hence, the required limit is \(\dfrac{19}{2}\).

Key Insight

Always check the denominator first. If it is non-zero at the given point, the limit becomes a straightforward substitution problem.

Exam Significance

• Very common in CBSE board exams (direct substitution case)
• Foundation for identifying indeterminate forms in JEE
• Helps distinguish between continuous vs undefined points

Concept Theory

A rational function \(\dfrac{p(x)}{q(x)}\) is continuous at a point if the denominator is non-zero there. In such cases, limits are evaluated using direct substitution.

Special care is required when dealing with powers of negative numbers:
Even powers → positive result, Odd powers → retain negative sign.

Solution Roadmap

• Check denominator at \(x = -1\)
• Evaluate powers carefully
• Substitute directly and simplify

Solution

Evaluate the denominator at \(x = -1\):

\[ x - 1 = -1 - 1 = -2 \neq 0 \]

Since the denominator is non-zero, the function is continuous at \(x = -1\). Therefore, we apply direct substitution:

\[ \begin{aligned} \lim_{x\rightarrow -1}\dfrac{x^{10}+x^{5}+1}{x-1} &= \dfrac{(-1)^{10}+(-1)^{5}+1}{-1-1} \ &= \dfrac{1 - 1 + 1}{-2} \ &= -\dfrac{1}{2} \end{aligned} \]

Hence, the required limit is \(-\dfrac{1}{2}\).

Key Insight

Always evaluate powers carefully when substituting negative values. Errors in sign handling are one of the most common mistakes in limits.

Exam Significance

• Tests accuracy in handling even and odd powers
• Common in CBSE exams for direct substitution with simplification
• Builds discipline for JEE algebraic manipulation

Concept Theory

When direct substitution gives the form \(\frac{0}{0}\), the limit is called an indeterminate form. In such cases, algebraic simplification is required.

A standard method is to use the Binomial Theorem:
\((1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots\)

These limits also represent the definition of derivative: \[ \lim_{x\to0}\frac{(1+x)^n - 1}{x} = n \]

Solution Roadmap

• Check form → \(0/0\)
• Expand using binomial theorem
• Cancel common factor \(x\)
• Substitute \(x = 0\)

Solution

On substituting \(x = 0\):

\[ \dfrac{(1+0)^5 - 1}{0} = \dfrac{1 - 1}{0} = \dfrac{0}{0} \]

Hence, we expand \((1+x)^5\):

\[ (1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 \]

Therefore,

\[ (1+x)^5 - 1 = 5x + 10x^2 + 10x^3 + 5x^4 + x^5 \]

Dividing by \(x\):

\[ \dfrac{(1+x)^5 - 1}{x} = 5 + 10x + 10x^2 + 5x^3 + x^4 \]

Now substitute \(x = 0\):

\[ \lim_{x\rightarrow 0} \left(5 + 10x + 10x^2 + 5x^3 + x^4\right) = 5 \]

Hence, the required limit is 5.

Key Insight

This is a standard result:
\(\lim_{x\to0} \frac{(1+x)^n - 1}{x} = n\).
It forms the foundation of derivative of \(x^n\).

Exam Significance

• Very important for JEE/NEET standard limits
• Directly linked to derivative definition
• Frequently appears in disguised forms in objective questions

Concept Theory

When direct substitution gives \(\frac{0}{0}\), the expression is indeterminate. This usually indicates the presence of a common factor that can be cancelled.

Such discontinuities are called removable discontinuities, where the function is undefined at a point but the limit exists.

Solution Roadmap

• Check form → \(0/0\)
• Factor numerator and denominator
• Cancel common factor
• Substitute the value

Solution

Substituting \(x = 2\):

\[ \begin{aligned} \dfrac{3(2)^2 - 2 - 10}{2^2 - 4} &= \dfrac{12 - 12}{0}\\ &= \dfrac{0}{0} \end{aligned} \]

Factor the numerator:

\[ \begin{aligned} 3x^2 - x - 10 &= 3x^2 - 6x + 5x - 10\\ &= (x-2)(3x+5) \end{aligned} \]

Factor the denominator:

\[ x^2 - 4 = (x-2)(x+2) \]

Cancel the common factor:

\[ \dfrac{(x-2)(3x+5)}{(x-2)(x+2)} = \dfrac{3x+5}{x+2} \]

Now substitute \(x = 2\):

\[ \dfrac{3(2)+5}{2+2} = \dfrac{11}{4} \]

Hence, the required limit is \(\dfrac{11}{4}\).

Key Insight

If a common factor cancels out, the discontinuity is removable and the limit depends on the simplified expression, not the original undefined form.

Exam Significance

• Very common JEE/NEET limit pattern
• Tests ability to detect hidden factors
• Foundation for understanding continuity and differentiability

Concept Theory

When substitution leads to \(\frac{0}{0}\), the expression must be simplified. This problem involves multiple factorization techniques:

• Difference of squares: \(a^2 - b^2 = (a-b)(a+b)\)
• Quadratic splitting (middle term factorization)

The goal is to identify and cancel the common factor causing the indeterminate form.

Solution Roadmap

• Check indeterminate form
• Factor numerator completely
• Factor denominator using splitting
• Cancel common factor
• Substitute the value

Solution

Substituting \(x = 3\):

\[ \begin{aligned} \dfrac{3^4 - 81}{2(3)^2 - 5(3) - 3} &= \dfrac{81 - 81}{18 - 15 - 3}\\ &= \dfrac{0}{0} \end{aligned} \]

Factor the numerator:

\[ \begin{aligned} x^4 - 81 &= (x^2 - 9)(x^2 + 9) \\&= (x-3)(x+3)(x^2+9) \end{aligned} \]

Factor the denominator:

\[ 2x^2 - 5x - 3 = (x-3)(2x+1) \]

Cancel the common factor \((x-3)\):

\[ \dfrac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} = \dfrac{(x+3)(x^2+9)}{2x+1} \]

Now substitute \(x = 3\):

\[ \begin{aligned} \dfrac{(3+3)(3^2+9)}{2(3)+1} &= \dfrac{6 \cdot 18}{7}\\ &= \dfrac{108}{7} \end{aligned} \]

Hence, the required limit is \(\dfrac{108}{7}\).

Key Insight

Complex expressions often hide simple cancellations. Always factor completely before substituting.

Exam Significance

• Tests mastery of multiple factorization techniques
• Very common in JEE Main & Advanced
• Strengthens algebra required for derivatives and continuity

Concept Theory

A rational function is continuous at a point if its denominator is non-zero there. Since \(cx+1 \neq 0\) at \(x=0\), the function is continuous and the limit can be evaluated using direct substitution.

This is a parameter-based limit, where constants \(a, b, c\) remain unchanged during evaluation.

Solution Roadmap

• Check denominator at \(x=0\)
• Apply direct substitution
• Simplify expression

Solution

Substitute \(x = 0\):

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{ax+b}{cx+1} &= \dfrac{a(0)+b}{c(0)+1} \\ &= \dfrac{b}{1} \\ &= b \end{aligned} \]

Hence, the required limit is \(b\).

Key Insight

In parameter-based limits, treat constants carefully and substitute only the variable.

Exam Significance

• Common in CBSE boards (1 mark direct)
• Foundation for JEE parameter-based limits
• Tests clarity in handling variables vs constants

Concept Theory

When substitution gives \(\frac{0}{0}\), we simplify using algebraic techniques. For fractional powers, a useful method is substitution to convert the expression into a standard algebraic form.

Expressions like \(\dfrac{u^2 - 1}{u - 1}\) are simplified using difference of squares.

Solution Roadmap

• Check indeterminate form
• Use substitution \(u = z^{1/6}\)
• Apply algebraic identity
• Substitute back and evaluate

Solution

Substituting \(z = 1\):

\[ \dfrac{1^{1/3} - 1}{1^{1/6} - 1} = \dfrac{0}{0} \]

Let \(u = z^{1/6}\). Then \(z^{1/3} = u^2\).

The expression becomes:

\[ \dfrac{u^2 - 1}{u - 1} \]

Factor:

\[ \dfrac{(u-1)(u+1)}{u-1} = u + 1 \]

Now as \(z \to 1\), \(u = z^{1/6} \to 1\).

\[ \lim_{z\rightarrow 1} (u+1) = 1 + 1 = 2 \]

Hence, the required limit is 2.

Key Insight

Substitution converts complex exponents into simple algebra, making factorization easier.

Exam Significance

• Very common in JEE (fractional exponent limits)
• Tests ability to use smart substitution
• Important for handling radical and power expressions

Concept Theory

A rational function is continuous at a point if its denominator is non-zero there. The given condition \(a+b+c \ne 0\) ensures that the denominator does not vanish at \(x=1\).

Such problems test whether the student correctly uses the given constraint before applying substitution.

Solution Roadmap

• Evaluate denominator at \(x=1\)
• Use given condition to confirm continuity
• Apply direct substitution
• Simplify carefully

Solution

Substitute \(x = 1\):

\[ \begin{aligned} \lim_{x\rightarrow 1}\dfrac{ax^2+bx+c}{cx^2+bx+a} &= \dfrac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} \\ &= \dfrac{a + b + c}{c + b + a} \end{aligned} \]

Rearranging the denominator:

\[ c + b + a = a + b + c \]

Hence,

\[ \dfrac{a+b+c}{a+b+c} = 1 \]

Since \(a+b+c \ne 0\), division is valid.

Therefore, the required limit is 1.

Key Insight

Always verify given conditions before simplifying. Without \(a+b+c \ne 0\), the expression could become undefined.

Exam Significance

• Frequently asked in CBSE boards with conditions
• Tests logical use of given constraints
• Important for JEE conceptual clarity in parameter-based limits

Concept Theory

When substitution gives \(\frac{0}{0}\), the expression must be simplified. In such cases, expressions involving fractions should be combined into a single rational expression.

The key idea is to convert the numerator into a single fraction so that common factors can be identified and cancelled.

Solution Roadmap

• Check indeterminate form
• Combine terms in numerator
• Factor and cancel common term
• Substitute the value

Solution

Substituting \(x = -2\):

\[ \dfrac{\frac{1}{-2} + \frac{1}{2}}{-2 + 2} = \dfrac{0}{0} \]

Combine the terms in the numerator:

\[ \frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x} \]

Substitute into the expression:

\[ \dfrac{\frac{2+x}{2x}}{x+2} = \dfrac{2+x}{2x(x+2)} \]

Since \(2 + x = x + 2\), we cancel the common factor:

\[ \dfrac{x+2}{2x(x+2)} = \dfrac{1}{2x} \]

Now substitute \(x = -2\):

\[ \dfrac{1}{2(-2)} = -\dfrac{1}{4} \]

Hence, the required limit is \(-\dfrac{1}{4}\).

Key Insight

Always combine fractional terms first. Hidden factors often appear only after simplification.

Exam Significance

• Tests skill in handling complex rational expressions
• Common in JEE algebra-based limits
• Strengthens manipulation needed for derivatives

Concept Theory

The standard trigonometric limit: \[ \lim_{t\to0}\frac{\sin t}{t} = 1 \] is fundamental in calculus.

To apply this, expressions must be transformed into the form \(\frac{\sin(\text{something})}{\text{same thing}}\).

Solution Roadmap

• Check indeterminate form
• Rewrite expression to match standard limit
• Apply identity \(\frac{\sin t}{t} \to 1\)
• Simplify constants

Solution

Substituting \(x = 0\):

\[ \dfrac{\sin(ax)}{bx} = \dfrac{0}{0} \]

Rewrite the expression:

\[ \dfrac{\sin ax}{bx} = \left(\dfrac{\sin ax}{ax}\right)\cdot \dfrac{a}{b} \]

As \(x \to 0\), we have \(ax \to 0\), so

\[ \dfrac{\sin ax}{ax} \to 1 \]

Therefore,

\[ \lim_{x\to0}\dfrac{\sin ax}{bx} = 1 \cdot \dfrac{a}{b} = \dfrac{a}{b} \]

Hence, the required limit is \(\dfrac{a}{b}\).

Key Insight

Always try to convert expressions into standard limit forms. This avoids unnecessary expansion.

Exam Significance

• One of the most important limits for JEE/NEET
• Used in derivation of derivatives of trigonometric functions
• Frequently appears in objective and subjective problems

Concept Theory

The standard limit: \[ \lim_{t\to0}\frac{\sin t}{t} = 1 \] is applied to both numerator and denominator.

For expressions like \(\frac{\sin(ax)}{\sin(bx)}\), we rewrite each term to match the standard form.

Solution Roadmap

• Check indeterminate form
• Introduce factors \(ax\) and \(bx\)
• Apply standard limit to both parts
• Simplify constants

Solution

Substituting \(x = 0\):

\[ \dfrac{\sin(ax)}{\sin(bx)} = \dfrac{0}{0} \]

Rewrite the expression:

\[ \dfrac{\sin ax}{\sin bx} = \left(\dfrac{\sin ax}{ax}\right) \cdot \dfrac{a}{b} \cdot \left(\dfrac{bx}{\sin bx}\right) \]

As \(x \to 0\), we have:

\[ \begin{aligned} \dfrac{\sin ax}{ax} \to 1 \\ \text{and} \\\ \dfrac{bx}{\sin bx} \to 1 \end{aligned} \]

Therefore,

\[ \begin{aligned} \lim_{x\to0}\dfrac{\sin ax}{\sin bx} &= 1 \cdot \dfrac{a}{b} \cdot 1\\ &= \dfrac{a}{b} \end{aligned} \]

Hence, the required limit is \(\dfrac{a}{b}\).

Key Insight

When both numerator and denominator behave like their angles near zero, their ratio reduces to the ratio of coefficients.

Exam Significance

• Core formula for JEE/NEET trigonometric limits
• Frequently appears in multi-step problems
• Essential for derivatives of trig functions

Concept Theory

When limits involve expressions like \((\pi - x)\), it is often useful to shift the variable so that the limit approaches 0, allowing the use of standard results.

Also, the identity: \[ \sin(\pi - x) = \sin x \] helps simplify trigonometric expressions.

Solution Roadmap

• Check indeterminate form
• Use identity \(\sin(\pi - x)\)
• Substitute \(t = \pi - x\)
• Apply standard limit \(\frac{\sin t}{t} \to 1\)

Solution

Substituting \(x = \pi\):

\[ \dfrac{\sin(\pi - \pi)}{\pi(\pi - \pi)} = \dfrac{0}{0} \]

Rewrite the expression:

\[ \dfrac{\sin(\pi - x)}{\pi(\pi - x)} = \dfrac{1}{\pi} \cdot \dfrac{\sin(\pi - x)}{\pi - x} \]

Let \(t = \pi - x\). Then as \(x \to \pi\), \(t \to 0\).

\[ \dfrac{1}{\pi} \lim_{t \to 0} \dfrac{\sin t}{t} \]

Using the standard limit:

\[ \lim_{t \to 0} \dfrac{\sin t}{t} = 1 \]

Therefore,

\[ \lim_{x\to\pi}\dfrac{\sin(\pi-x)}{\pi(\pi-x)} = \dfrac{1}{\pi} \]

Hence, the required limit is \(\dfrac{1}{\pi}\).

Key Insight

Shifting the variable to make the limit approach zero is a powerful strategy for applying standard trigonometric limits.

Exam Significance

• Very common in JEE/NEET as shifted limits
• Tests understanding of trigonometric identities + substitution
• Foundation for advanced limit transformations

Concept Theory

The function \(\cos x\) is continuous for all real values of \(x\). A rational function is continuous at a point if its denominator is non-zero there.

If no indeterminate form arises, the limit can be evaluated using direct substitution.

Solution Roadmap

• Check denominator at \(x = 0\)
• Confirm continuity
• Apply direct substitution

Solution

Substitute \(x = 0\):

\[ \lim_{x\rightarrow 0}\dfrac{\cos x}{\pi - x} = \dfrac{\cos 0}{\pi - 0} \]

Since \(\cos 0 = 1\), we get:

\[ \dfrac{1}{\pi} \]

Hence, the required limit is \(\dfrac{1}{\pi}\).

Key Insight

Not all trigonometric limits require identities. Always check for continuity first before applying advanced techniques.

Exam Significance

• Tests ability to identify non-indeterminate cases
• Common in CBSE exams to trap overthinking
• Builds decision-making for JEE problem-solving

Concept Theory

When trigonometric expressions lead to \(\frac{0}{0}\), we simplify using identities.

A key identity used here is: \[ \cos 2x = 2\cos^2 x - 1 \]

This helps convert the numerator into a form involving \((\cos x - 1)\), allowing cancellation.

Solution Roadmap

• Check indeterminate form
• Use identity for \(\cos 2x\)
• Factor expression
• Cancel common term
• Substitute \(x=0\)

Solution

Substituting \(x = 0\):

\[ \dfrac{\cos 0 - 1}{\cos 0 - 1} = \dfrac{0}{0} \]

Use identity:

\[ \begin{aligned} \cos 2x - 1 &= 2\cos^2 x - 1 - 1 \\&= 2(\cos^2 x - 1) \end{aligned} \]

So the expression becomes:

\[ \dfrac{2(\cos^2 x - 1)}{\cos x - 1} \]

Factor:

\[ \cos^2 x - 1 = (\cos x - 1)(\cos x + 1) \]

Cancel common factor:

\[ \dfrac{2(\cos x - 1)(\cos x + 1)}{\cos x - 1} = 2(\cos x + 1) \]

Now substitute \(x = 0\):

\[ 2(1 + 1) = 4 \]

Hence, the required limit is 4.

Key Insight

Choosing the right identity simplifies the problem drastically. Avoid unnecessary manipulation when a direct identity works.

Exam Significance

• Common in JEE/NEET identity-based limits
• Tests ability to pick the most efficient identity
• Important for trigonometric simplification skills

Concept Theory

When expressions involve both algebraic and trigonometric terms, we combine:

• Factoring common terms
• Standard limit: \(\lim_{x\to0}\frac{\sin x}{x} = 1\)
• Continuity of \(\cos x\)

The goal is to rewrite the expression so that standard limits can be applied directly.

Solution Roadmap

• Check indeterminate form
• Factor \(x\) from numerator
• Introduce \(\frac{\sin x}{x}\)
• Apply limits

Solution

Substituting \(x = 0\):

\[ \dfrac{0 + 0}{0} = \dfrac{0}{0} \]

Factor \(x\) from the numerator:

\[ ax + x\cos x = x(a + \cos x) \]

So the expression becomes:

\[ \dfrac{x(a + \cos x)}{b\sin x} \]

Rewrite using standard limit form:

\[ \dfrac{a + \cos x}{b} \cdot \dfrac{x}{\sin x} \]

Now as \(x \to 0\):

\[ \begin{aligned} \cos x \to 1 \\ \text{and} \\ \dfrac{\sin x}{x} \to 1 \\\Rightarrow \dfrac{x}{\sin x} \to 1 \end{aligned} \]

Therefore,

\[ \lim_{x\to0}\dfrac{ax+x\cos x}{b\sin x} = \dfrac{a + 1}{b} \]

Hence, the required limit is \(\dfrac{a+1}{b}\).

Key Insight

Always try to introduce \(\frac{\sin x}{x}\) or its reciprocal. It is the backbone of most trigonometric limits.

Exam Significance

• Very common in JEE mixed limits
• Tests ability to combine algebra + trigonometry
• Important for derivative-based limits

Concept Theory

The function \(\sec x = \dfrac{1}{\cos x}\) is continuous near \(x=0\), and \(\cos 0 = 1 \neq 0\), so \(\sec x\) remains finite (bounded) near zero.

A key idea:
If one factor → 0 and the other remains bounded, then the product → 0.

Solution Roadmap

• Rewrite \(\sec x\)
• Check behavior of numerator and denominator
• Apply direct substitution

Solution

Rewrite:

\[ x\sec x = \frac{x}{\cos x} \]

As \(x \to 0\):

\[ x \to 0 \quad \text{and} \quad \cos x \to 1 \]

Therefore,

\[ \begin{aligned} \lim\limits_{x\to 0} \frac{x}{\cos x} &= \frac{0}{1}\\ &= 0 \end{aligned} \]

Hence, the required limit is 0.

Key Insight

Whenever a small quantity is multiplied by a bounded function, the result tends to zero.

Exam Significance

• Tests understanding of limit behavior (bounded × 0)
• Common shortcut in JEE objective questions
• Avoids unnecessary use of standard limits

Concept Theory

For limits involving \(\sin(ax)\) and \(\sin(bx)\), we use: \[ \lim_{x\to0}\frac{\sin kx}{x} = k \]

Dividing numerator and denominator by \(x\) simplifies the expression and allows direct use of standard limits.

Solution Roadmap

• Check indeterminate form
• Divide numerator and denominator by \(x\)
• Apply standard limits
• Simplify

Solution

Substituting \(x = 0\):

\[ \dfrac{0 + 0}{0 + 0} = \dfrac{0}{0} \]

Divide numerator and denominator by \(x\):

\[ \dfrac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}} \]

Using standard limits:

\[ \begin{aligned} \frac{\sin ax}{x} &\to a \quad \text{and} \\ \frac{\sin bx}{x} &\to b \end{aligned} \]

Therefore,

\[ \lim_{x\to0}\dfrac{\sin ax+bx}{ax+\sin bx} = \dfrac{a + b}{a + b} = 1 \]

Since \(a + b \ne 0\), division is valid.

Hence, the required limit is 1.

Key Insight

Symmetric expressions in numerator and denominator often simplify to 1. Always look for balance before doing lengthy manipulation.

Exam Significance

• Very common JEE symmetric limit pattern
• Tests ability to simplify using standard limits efficiently
• Helps avoid long algebra in objective exams

Concept Theory

Expressions involving \(\text{cosec }x\) and \(\cot x\) are often simplified by converting them into sine and cosine:

\[ \begin{aligned} \text{cosec }x &= \frac{1}{\sin x}, \\ \cot x &= \frac{\cos x}{\sin x} \end{aligned} \]

This helps combine them into a single fraction, revealing hidden identities.

A key identity used: \[ 1 - \cos x = 2\sin^2\frac{x}{2} \]

Solution Roadmap

• Rewrite in sine and cosine
• Combine into single fraction
• Apply identity \(1 - \cos x\)
• Simplify and substitute

Solution

Rewrite:

\[ \begin{aligned} \text{cosec }x - \cot x &= \frac{1}{\sin x} - \frac{\cos x}{\sin x}\\ &= \frac{1 - \cos x}{\sin x} \end{aligned} \]

Substituting \(x = 0\) gives \(\frac{0}{0}\), so we simplify.

Use identity:

\[ \begin{aligned} 1 - \cos x &= 2\sin^2\frac{x}{2} \quad \text{and} \\ \sin x &= 2\sin\frac{x}{2}\cos\frac{x}{2} \end{aligned} \]

Substitute:

\[ \begin{aligned} \frac{1 - \cos x}{\sin x} &= \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\\ &= \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \end{aligned} \]

Now take limit:

\[ \ \begin{aligned} \lim_{x\to0} \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} &= \frac{0}{1} \\&= 0 \end{aligned} \]

Hence, the required limit is 0.

Key Insight

Expressions of the form \(\frac{1 - \cos x}{\sin x}\) are best handled using half-angle identities.

Exam Significance

• Very common JEE identity-based limit
• Tests ability to use half-angle identities
• Builds skill for simplifying trigonometric expressions

Concept Theory

When limits involve expressions like \(x - \frac{\pi}{2}\), shifting the variable helps convert the limit into a standard form near 0.

Also, tangent has the periodic property: \[ \tan(\theta + \pi) = \tan \theta \]

Standard limit used: \[ \lim_{t\to0}\frac{\tan t}{t} = 1 \]

Solution Roadmap

• Shift variable: \(y = x - \frac{\pi}{2}\)
• Use periodicity of tangent
• Convert into standard limit form
• Evaluate

Solution

Let \(y = x - \frac{\pi}{2}\). Then as \(x \to \frac{\pi}{2}\), \(y \to 0\).

\[ x = y + \frac{\pi}{2} \]

Substitute into the expression:

\[ \lim_{y\to0} \frac{\tan\left(2y + \pi\right)}{y} \]

Using periodicity:

\[ \tan(2y + \pi) = \tan(2y) \]

So the limit becomes:

\[ \lim_{y\to0} \frac{\tan 2y}{y} \]

Rewrite:

\[ \frac{\tan 2y}{y} = 2 \cdot \frac{\tan 2y}{2y} \]

Using standard limit:

\[ \lim_{y\to0}\frac{\tan 2y}{2y} = 1 \]

Therefore,

\[ \lim_{x\to\frac{\pi}{2}}\dfrac{\tan 2x}{x-\frac{\pi}{2}} = 2 \]

Hence, the required limit is 2.

Key Insight

Shifting the variable and using periodicity reduces complex limits to standard forms.

Exam Significance

• Classic JEE Advanced level transformation
• Tests understanding of periodicity + substitution
• Important for mastering non-zero approach limits

Concept Theory

For piecewise functions, limits at a point where the definition changes must be evaluated using:

• Left-Hand Limit (LHL): approaching from left
• Right-Hand Limit (RHL): approaching from right

The limit exists only if: \[ \text{LHL} = \text{RHL} \]

Solution Roadmap

• Check point where definition changes (\(x=0\))
• Find LHL and RHL separately
• Compare values
• Evaluate second limit normally

Solution

Limit as \(x \to 0\)

Left-hand limit:

\[ \begin{aligned} \lim_{x \to 0^-} f(x) &= \lim_{x \to 0^-} (2x + 3) \\&= 3 \end{aligned} \]

Right-hand limit:

\[ \begin{aligned} \lim_{x \to 0^+} f(x) &= \lim_{x \to 0^+} 3(x+1) \\&= 3 \end{aligned} \]

Since LHL = RHL,

\[ \lim_{x \to 0} f(x) = 3 \]

Limit as \(x \to 1\)

For values near \(x=1\), we use \(x > 0\), so:

\[ f(x) = 3(x+1) \]

\[ \begin{aligned} \lim_{x \to 1} f(x) &= 3(1+1) \\&= 6 \end{aligned} \]

Hence,
\(\lim_{x \to 0} f(x) = 3\) and \(\lim_{x \to 1} f(x) = 6\).

Key Insight

Always check both sides at the point where the function changes its definition.

Exam Significance

• Very important for CBSE board exams (piecewise functions)
• Foundation for continuity and differentiability
• Frequently used in JEE conceptual questions

Concept Theory

For piecewise functions, checking LHL and RHL is required only if the point lies at the boundary where the function changes definition.

If the point lies entirely within one interval, the function behaves like a normal continuous function in that region.

Solution Roadmap

• Identify region of the point \(x=1\)
• Select correct expression
• Apply direct substitution

Solution

Since \(1 > 0\), all values of \(x\) near 1 satisfy \(x > 0\).

Therefore, we use:

\[ f(x) = -x^2 - 1 \]

Now evaluate the limit:

\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} (-x^2 - 1) \]

\[ = -(1)^2 - 1 = -2 \]

Hence, the required limit is \(-2\).

Key Insight

Always check whether the point lies at the boundary. If not, avoid unnecessary LHL/RHL work.

Exam Significance

• Common trick in CBSE board exams
• Tests understanding of piecewise intervals
• Important for continuity problems in JEE

Concept Theory

The expression \(\dfrac{|x|}{x}\) depends on the sign of \(x\):

\[ \frac{|x|}{x} = \begin{cases} -1, & x < 0 \ 1, & x > 0 \end{cases} \]

For limits at such points, we must evaluate: Left-Hand Limit (LHL) and Right-Hand Limit (RHL).

Solution Roadmap

• Analyze expression for \(x < 0\) and \(x > 0\)
• Find LHL and RHL
• Compare values

Solution

For \(x < 0\), \(|x| = -x\), so:

\[ \lim_{x \to 0^-} \frac{|x|}{x} = \frac{-x}{x} = -1 \]

For \(x > 0\), \(|x| = x\), so:

\[ \lim_{x \to 0^+} \frac{|x|}{x} = \frac{x}{x} = 1 \]

Since:

\[ \lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x) \]

Therefore, the limit does not exist.

Note: \(f(0) = 0\), but the limit depends only on nearby values, not the function value at the point.

Key Insight

If LHL ≠ RHL, the limit does not exist, even if the function is defined at that point.

Exam Significance

• Classic example of non-existent limit
• Very important for continuity concepts
• Frequently asked in CBSE and JEE

Concept Theory

The expression \(\dfrac{x}{|x|}\) depends on the sign of \(x\):

\[ \frac{x}{|x|} = \begin{cases} -1, & x < 0 \ 1, & x > 0 \end{cases} \]

This is known as a sign-based (signum-type) function. Limits must be evaluated using LHL and RHL.

Solution Roadmap

• Evaluate for \(x < 0\) and \(x > 0\)
• Find LHL and RHL
• Compare values

Solution

For \(x < 0\), \(|x| = -x\):

\[ \begin{aligned} \lim_{x \to 0^-} \frac{x}{|x|} &= \frac{x}{-x} \\&= -1 \end{aligned} \]

For \(x > 0\), \(|x| = x\):

\[ \begin{aligned} \lim_{x \to 0^+} \frac{x}{|x|} &= \frac{x}{x} \\&= 1 \end{aligned} \]

Since:

\[ \lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x) \]

Therefore, the limit does not exist.

Note: \(f(0) = 0\), but this does not affect the existence of the limit.

Key Insight

Expressions like \(\frac{x}{|x|}\) behave like step functions, causing jump discontinuity.

Exam Significance

• Very common example of non-existent limit
• Important for understanding signum and modulus functions
• Used in JEE conceptual questions

Concept Theory

The modulus function \(|x|\) is continuous for all real values of \(x\). Therefore, \(f(x) = |x| - 5\) is also continuous everywhere.

For continuous functions:
Limit = Function value

Solution Roadmap

• Check continuity
• Apply direct substitution

Solution

Since \(f(x)\) is continuous at \(x = 5\),

\[ \begin{aligned} \lim_{x \to 5} (|x| - 5) &= |5| - 5 \\&= 5 - 5 \\&= 0 \end{aligned} \]

Hence, the required limit is 0.

Key Insight

Modulus functions are continuous everywhere, so limits are straightforward.

Exam Significance

• Direct question in CBSE boards
• Helps identify when no LHL/RHL needed

Concept Theory

The condition \(\lim_{x\to1} f(x) = f(1)\) means the function is continuous at \(x=1\).

For continuity: \[ \text{LHL} = \text{RHL} = f(1) \]

Solution Roadmap

• Find LHL using left expression
• Find RHL using right expression
• Equate both to \(f(1)\)
• Solve simultaneous equations

Solution

Left-hand limit:

\[ \lim_{x \to 1^-} f(x) = a + b \]

Right-hand limit:

\[ \lim_{x \to 1^+} f(x) = b - a \]

Given:

\[ f(1) = 4 \]

Apply continuity:

\[ a + b = 4 \tag{1} \] \[ b - a = 4 \tag{2} \]

Add (1) and (2):

\[ \begin{aligned} 2b &= 8 \\\Rightarrow b &= 4 \end{aligned} \]

Substitute in (1):

\[ \begin{aligned} a + 4 &= 4 \\\Rightarrow a &= 0 \end{aligned} \]

Hence, the required values are:
\(a = 0,\; b = 4\).

Key Insight

Continuity problems reduce to solving equations obtained from LHL = RHL = value.

Exam Significance

• Very important for CBSE (3–5 marks)
• Core concept in JEE continuity problems
• Foundation for differentiability

Concept Theory

The given function is a polynomial, which is continuous for all real values of \(x\).

Important idea:

• If \(x = a_i\), one factor becomes zero → entire product becomes zero
• If \(x \ne a_i\), evaluate directly using substitution

Solution Roadmap

• Use continuity of polynomial
• Substitute \(x = a_1\)
• Substitute \(x = a\)

Solution

Case 1: \(x \to a_1\)

\[ \lim_{x \to a_1} (x-a_1)(x-a_2)\ldots(x-a_n) \]

Substituting \(x = a_1\):

\[ \begin{aligned} &= (a_1-a_1)(a_1-a_2)\ldots(a_1-a_n)\\ &= 0 \end{aligned} \]

Hence,
\(\lim_{x \to a_1} f(x) = 0\)

Case 2: \(x \to a\), where \(a \ne a_1, a_2, \ldots, a_n\)

Since no factor becomes zero, substitute directly:

\[ \lim_{x \to a} f(x) = (a-a_1)(a-a_2)\ldots(a-a_n) \]

Hence,
\(\lim_{x \to a} f(x) = (a-a_1)(a-a_2)\ldots(a-a_n)\)

Key Insight

A polynomial becomes zero at its roots. Limits at roots directly reflect this property.

Exam Significance

• Important for understanding roots of polynomials
• Used in JEE polynomial-based limits
• Foundation for factor theorem and continuity

Concept Theory

The limit at a point exists if the function approaches the same value from both sides.

For piecewise functions:

• Inside an interval → function behaves continuously
• At boundary → check LHL and RHL

Solution Roadmap

• Check intervals \(a < 0\) and \(a > 0\)
• Analyze boundary point \(a = 0\)
• Compare LHL and RHL

Solution

Case 1: \(a < 0\)

Here, \(f(x) = |x| + 1\), which is continuous.
Hence, limit exists for all \(a < 0\).

Case 2: \(a > 0\)

Here, \(f(x) = |x| - 1\), also continuous.
Hence, limit exists for all \(a > 0\).

Case 3: \(a = 0\)

Left-hand limit:

\[ \lim_{x \to 0^-} (|x| + 1) = 1 \]

Right-hand limit:

\[ \lim_{x \to 0^+} (|x| - 1) = -1 \]

Since:

\[ \lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x) \]

Therefore, the limit does not exist at \(x = 0\).

Hence, the limit exists for:
all real \(a \ne 0\).

Key Insight

Limits fail to exist only at points where the function shows a jump (LHL ≠ RHL).

Exam Significance

• Important for existence of limits
• Common in CBSE 3–5 mark questions
• Core concept for continuity classification (jump discontinuity)

Concept Theory

If a limit of the form \(\dfrac{\text{something}}{0}\) is finite, then the numerator must also approach 0.

This is based on the idea:
If denominator → 0 and limit is finite ⇒ numerator → 0.

Solution Roadmap

• Analyze denominator as \(x \to 1\)
• Use condition of finite limit
• Deduce value of \(\lim f(x)\)
• Check information for second limit

Solution

Given:

\[ \lim_{x \to 1} \dfrac{f(x)-2}{x^2 - 1} = \pi \]

As \(x \to 1\),

\[ x^2 - 1 \to 0 \]

Since the limit is finite (\(\pi\)), the numerator must approach zero:

\[ \lim_{x \to 1} (f(x) - 2) = 0 \]

Therefore,

\[ \lim_{x \to 1} f(x) = 2 \]

Hence, the limit at \(x = 1\) exists and equals 2.

Now consider \(\lim_{x \to 0} f(x)\)

The given condition provides information only near \(x = 1\). No information is given about the behavior of \(f(x)\) near \(x = 0\).

Therefore,

\(\lim_{x \to 0} f(x)\) cannot be determined.

Key Insight

A finite limit of a quotient forces both numerator and denominator to approach zero.

Exam Significance

• Very important for JEE conceptual problems
• Used in derivative definition
• Tests logical reasoning beyond computation

Concept Theory

For a limit to exist at a point:

• LHL = RHL
• Function value is not required for existence

For piecewise functions, only boundary points need checking.

Solution Roadmap

• Check limit at \(x=0\)
• Check limit at \(x=1\)
• Form equations
• Find condition on \(m,n\)

Solution

At \(x = 0\)

Left-hand limit:

\[ \lim_{x \to 0^-} (mx^2 + n) = n \]

Right-hand limit:

\[ \lim_{x \to 0^+} (nx + m) = m \]

For limit to exist:

\[ n = m \quad ...(1) \]

At \(x = 1\)

Left-hand limit:

\[ \lim_{x \to 1^-} (nx + m) = n + m \]

Right-hand limit:

\[ \lim_{x \to 1^+} (nx^3 + m) = n + m \]

Since both are equal automatically, the limit at \(x = 1\) exists for all \(m,n\).

Hence, the only condition required is:

\(m = n\)

Therefore, for all integers satisfying \(m = n\), both limits exist.

Key Insight

Not all boundary points give conditions. Always check whether equality is automatic.

Exam Significance

• High-value question in CBSE (case-based)
• Tests ability to identify necessary vs redundant conditions
• Core foundation for continuity and differentiability