In mathematics, situations often arise where a variable quantity changes continuously and approaches a certain value without actually attaining it. To study such behavior rigorously, the concept of a limit is introduced. Consider a function \(f(x)\). When the values of \(x\) are taken closer and closer to a fixed real number \(a\), the corresponding values of \(f(x)\) may approach a definite real number \(L\). This number \(L\) is called the limit of the function as \(x\) approaches \(a\).

The idea of a limit depends on the behavior of the function in the neighborhood of the point and not necessarily on the value of the function at the point itself. Thus, even if \(f(a)\) is not defined, the limit \(\lim_{x\to a} f(x)\) may still exist.

Formally, let \(f(x)\) be a function defined in some interval around a real number \(a\), except possibly at \(x=a\). If the values of \(f(x)\) approach a unique real number \(L\) as \(x\) approaches \(a\), then \(L\) is called the limit of \(f(x)\) as \(x\) approaches \(a\), and we write \[ \lim_{x\to a} f(x)=L. \]

The approach of xtowards acan occur from two directions. If xapproaches athrough values smaller than a, the limit is called the left-hand limit and is denoted by \[{\lim}\limits_{x\rightarrow a^-}f(x)\] If xapproaches athrough values greater than a, the limit is called the right-hand limit and is denoted by \[{\lim}\limits_{x\rightarrow a^+}f(x)\] The limit of a function at x=aexists if and only if both the left-hand and right-hand limits exist and are equal.

A function is said to have a limit at x=awhen the following condition holds: \[{\lim}\limits_{x\rightarrow a^-}f(x)={\lim}\limits_{x\rightarrow a^+}f(x)\] If the two one-sided limits are unequal, then the limit of the function at that point does not exist. This condition emphasizes that the function must approach the same value from both sides of the point.

If the limits of functions \(f(x)\) and \(g(x)\) exist as \(x\rightarrow a\), then the following results hold: \[ \begin{aligned} \lim_{x\to a}[f(x)+g(x)] &= \lim\limits_{x \to a}f(x) + \lim\limits_{x \to a}g(x) \\\\ \lim_{x\to a}[f(x)-g(x)] &= \lim\limits_{x \to a}f(x) - \lim\limits_{x \to a}g(x) \\\\ \lim_{x\to a}[k f(x)] &= k\lim\limits_{x \to a}f(x) \\\\ \lim_{x\to a}[f(x)g(x)] &= (\lim\limits_{x \to a}f(x)) + (\lim\limits_{x \to a}g(x)) \\\\ \lim_{x\to a}\frac{f(x)}{g(x)} &= \dfrac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}, \quad \lim\limits_{x \to a}g(x)\neq 0. \end{aligned} \]

For polynomial functions, the limit at any real number can be found by direct substitution. That is, if \(f(x)\) is a polynomial, then \[ \lim_{x\to a} f(x)=f(a). \] Similarly, for a rational function \(\frac{p(x)}{q(x)}\), if \(q(a)\neq 0\), the limit is obtained by substituting \(x=a\).

Sometimes direct substitution leads to expressions such as \(\frac{0}{0}\), which are called indeterminate forms. In such cases, algebraic simplification is required before evaluating the limit. This may involve factorization, cancellation of common terms, or rationalization. After simplification, the limit can usually be evaluated by substitution.

Certain trigonometric limits occur frequently and play a fundamental role in calculus. Some important results are \[ \begin{aligned} &\lim_{x\to 0} \frac{\sin x}{x} = 1, \\\\ &\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}, \\\\ &\lim_{x\to 0} \frac{\tan x}{x} = 1. \end{aligned} \] These limits are essential for evaluating trigonometric expressions and for defining derivatives of trigonometric functions.

If the value of f(x)increases or decreases without bound as xapproaches a particular value, the limit is said to be infinite. For example, \[\lim_{x \to 0^+}\frac{1}{x}=\infty\] This indicates that the function grows indefinitely as xapproaches zero from the right.

The concept of limits is closely related to continuity. A function is said to be continuous at a point \(x=a\) if the following three conditions are satisfied simultaneously:

1. \[f(a) \text{ is defined}\]
2. \[\lim_{x\to a} f(x) \text{ exists}\]
3. \[\lim_{x\to a} f(x)=f(a)\]

Thus, continuity requires the existence of a limit, but the existence of a limit alone does not ensure continuity.

Significance of Limits

The concept of limits forms the foundation of calculus. It is essential for defining continuity, differentiation, and integration. Limits enable the study of instantaneous rates of change and provide mathematical tools to analyze physical, economic, and scientific processes involving gradual variation.

A strong understanding of limits is therefore crucial for mastering higher mathematics.

In many real-life situations, we are interested not only in how much a quantity changes, but also in how fast it changes. For example, when a body moves along a straight line, its position changes with time. The speed of the body at a particular instant cannot be determined by average speed over a large interval of time. What is required is a measure of change at an instant.

Mathematically, this leads to the study of the rate of change of one quantity with respect to another. The concept of the derivative provides a precise method for studying such instantaneous change.

The derivative of a function f(x)at a point xis defined as the limit of the average rate of change as the increment approaches zero.

Mathematically, the derivative of f(x)is defined by \[f^\prime(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\] provided this limit exists. This definition is known as differentiation from first principles.

Let \(f(x)=c\), where \(c\) is a constant.
Using the definition of derivative,
\[ \begin{aligned} f^\prime(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to 0}\frac{c-c}{h}\\ &=\lim_{h\to 0}0\\ &=0 \end{aligned} \] Thus, the derivative of a constant function is zero. This shows that a constant quantity does not change.

Let f(x)=x. Then, \[ \begin{aligned} f^\prime(x)&=\lim_{h\to 0}\frac{(x+h)-x}{h}\\ &=\lim_{h\to 0}\frac{h}{h}\\ &=\lim_{h\to0}1\\ &=1 \end{aligned} \] Hence, the derivative of \(x\) with respect to \(x\) is 1.

Let \(f(x)=x^n\), where \(n\) is a positive integer.
Using the definition, \[ \begin{aligned} f^\prime(x)&=\lim_{h\to 0}\frac{\left(x+h)^n-x^n\right.}{h} \end{aligned} \] Expanding \((x+h)^n\) using the binomial theorem, \[ \left(x+h)^n=x^n+nx^{n-1}h+\mathrm{terms\ containing\ } h^2.\right. \] Substituting, \[ \begin{aligned} f'(x)&= \lim_{h\to0}\dfrac{nx^{n-1}h+\text{higher power of h}}{h}\\\\ &= \lim_{h\to 0}(nx^{x-1} + \text{terms containing h})\\\\ &=nx^{n-1} \end{aligned} \] Thus, \[\boxed{\bbox[blue]{\;\;\dfrac{d}{dx}(x^n)=nx^{n-1}\;\;}}\]

Using limits and standard trigonometric identities, it can be shown that

• \[\frac{d}{dx}(\sin x)=\cos x\]
• \[\frac{d}{dx}(\cos x)=-\sin x\]

These results rely on fundamental trigonometric limits and form the basis for differentiation of other trigonometric expressions.

If \(f(x)\) and \(g(x)\) are differentiable functions and kis a constant, then:

• \[\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)\]
• \[\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)\]
• \[\frac{d}{dx}[kf(x)]=kf'(x)\]

These properties allow complex functions to be differentiated by breaking them into simpler components.

If \(s(t)\)denotes the position of a particle at time \(t\), then the derivative \(\frac{ds}{dt}\) represents the velocity of the particle. Similarly, the derivative of velocity with respect to time represents acceleration.
Thus, derivatives provide a mathematical framework for describing motion and change in physical systems.

If a function is differentiable at a point, then it must be continuous at that point. However, the converse is not always true. A function may be continuous at a point but not differentiable there, such as at sharp corners or points of abrupt change.

Find the limits: \(\lim\limits_{x\to0} [x^3-x^2+1]\)

Solution

Since the given expression is a polynomial, it is continuous for all real values of x. Therefore, the limit as x approaches 0 can be found by directly substituting x = 0 into the expression.

$$ \begin{aligned} \lim_{x \to 0} \left( x^{3} - x^{2} + 1 \right) \\ = 0^{3} - 0^{2} + 1 \\ = 1 \end{aligned} $$

Hence, the required limit is equal to 1.

Find the limits: \(\lim\limits_{x\to2}\left[\dfrac{x^2-4}{x^3-4x^2+4x}\right]\)

Solution

\[ \begin{aligned} \lim_{x\to 2}\left[\dfrac{x^{2}-4}{x^{3}-4x^{2}+4x}\right] \end{aligned} \]

First, we check the value of the function at \(x=2\) \[ \begin{aligned} \dfrac{2^{2}-4}{2^{3}-4\cdot 2^{2}+4\cdot 2} = \dfrac{4-4}{8-16+8} = \dfrac{0}{0} \end{aligned} \]

Since the limit gives the indeterminate form \(\dfrac{0}{0}\), we simplify the rational expression by factorisation \[ \begin{aligned} \lim_{x\to 2}\dfrac{x^{2}-4}{x^{3}-4x^{2}+4x} = \lim_{x\to 2}\dfrac{(x-2)(x+2)}{x(x^{2}-4x+4)} \end{aligned} \]

Factorising completely \[ \begin{aligned} \lim_{x\to 2}\dfrac{(x-2)(x+2)}{x(x-2)^{2}} \end{aligned} \]

Cancelling one common factor \((x-2)\) \[ \begin{aligned} \lim_{x\to 2}\dfrac{x+2}{x(x-2)} \end{aligned} \]

Now substitute \(x=2\) \[ \begin{aligned} \dfrac{2+2}{2(2-2)}=\dfrac{4}{0} \end{aligned} \]

The denominator becomes zero while the numerator remains finite, hence the expression grows without bound as \(x\) approaches \(2\). Therefore, the given limit does not exist and is not defined at \(x=2\).

Find the limits: \(\lim\limits_{x\to1}\left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]\)

Solution

\[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{x-2}{x^{2}-x}-\dfrac{1}{x^{3}-3x^{2}+2x}\right] \end{aligned} \]

First, substitute \(x=1\) to examine the nature of the limit \[ \begin{aligned} \dfrac{1-2}{1^{2}-1}-\dfrac{1}{1^{3}-3\cdot1^{2}+2\cdot1} = \dfrac{-1}{0}-\dfrac{1}{0} \end{aligned} \]

Since both terms are undefined at \(x=1\), the given expression is of indeterminate type and must be simplified by factorisation \[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x^{2}-3x+2)}\right] \end{aligned} \]

Factorising the quadratic term in the second denominator \[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x-1)(x-2)}\right] \end{aligned} \]

Taking the LCM of the two fractions \[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{(x-2)^{2}-1}{x(x-1)(x-2)}\right] \end{aligned} \]

Simplifying the numerator \[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{x^{2}-4x+4-1}{x(x-1)(x-2)}\right] = \lim_{x\to 1}\left[\dfrac{x^{2}-4x+3}{x(x-1)(x-2)}\right] \end{aligned} \]

Factorising the quadratic expression in the numerator \[ \begin{aligned} \lim_{x\to 1}\left[\dfrac{(x-1)(x-3)}{x(x-1)(x-2)}\right] \end{aligned} \]

Cancelling the common factor \((x-1)\) \[ \begin{aligned} \lim_{x\to 1}\dfrac{x-3}{x(x-2)} \end{aligned} \]

Now substituting \(x=1\) \[ \begin{aligned} \dfrac{1-3}{1(1-2)}=\dfrac{-2}{-1}=2 \end{aligned} \]

Hence, the value of the given limit is \(2\).

Evaluate: \(\lim\limits_{x\to1}\dfrac{x^{15}-1}{x^{10}-1}\)

Solution

We first observe that direct substitution of \(x=1\) gives an indeterminate form

$$ \begin{aligned} \lim_{x\to 1}\frac{x^{15}-1}{x^{10}-1} = \frac{1^{15}-1}{1^{10}-1} = \frac{0}{0} \end{aligned} $$

To remove this indeterminacy, rewrite the given expression by expressing both powers in terms of \(x^{5}\)

$$ \begin{aligned} \frac{x^{15}-1}{x^{10}-1} = \frac{(x^{5})^{3}-1}{(x^{5})^{2}-1} \end{aligned} $$

Let \(u = x^{5}\). Then the expression becomes

$$ \begin{aligned} \frac{u^{3}-1}{u^{2}-1} &= \frac{(u-1)(u^{2}+u+1)}{(u-1)(u+1)}\\\\ &= \frac{u^{2}+u+1}{u+1} \end{aligned} $$

Substituting back \(u = x^{5}\) and now applying the limit, which is valid since the expression is no longer indeterminate

$$ \begin{aligned} &\lim_{x\to 1}\frac{(x^{5})^{2}+x^{5}+1}{x^{5}+1}\\\\ &= \frac{1^{2}+1+1}{1+1}\\\\ &= \frac{3}{2} \end{aligned} $$

Hence, the value of the given limit is \(\dfrac{3}{2}\)

Evaluate: \(\lim\limits_{x\to0}\dfrac{\sqrt{1+x}-1}{x}\)

Solution

On substituting \(x=0\) directly in the given expression, we obtain

$$ \begin{aligned} \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} = \frac{\sqrt{1+0}-1}{0} = \frac{0}{0} \end{aligned} $$

Thus, the limit is of the indeterminate form \(\frac{0}{0}\) and needs further simplification

$$ \begin{aligned} &\lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}\\\\ &= \lim_{x\to 0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} \end{aligned} $$

Multiplying numerator and denominator by the conjugate helps to rationalize the expression

$$ \begin{aligned} &= \lim_{x\to 0}\frac{(\sqrt{1+x})^{2}-1}{x(\sqrt{1+x}+1)}\\\\ &= \lim_{x\to 0}\frac{1+x-1}{x(\sqrt{1+x}+1)} \end{aligned} $$

Now cancelling the common factor \(x\) from the numerator and denominator

$$ \begin{aligned} &= \lim_{x\to 0}\frac{1}{\sqrt{1+x}+1}\\\\ &= \frac{1}{\sqrt{1+0}+1}\\\\ &= \frac{1}{2} \end{aligned} $$

Hence, the value of the given limit is \(\dfrac{1}{2}\)

Evaluate: \(\lim\limits_{x\to0}\dfrac{\sin 4x}{\sin 2x}\)

Solution

By direct substitution of \(x=0\) in the given expression, we obtain

$$ \begin{aligned} \lim_{x\to 0}\frac{\sin 4x}{\sin 2x} &= \frac{\sin 0}{\sin 0}\\ &= \frac{0}{0} \end{aligned} $$

Since the limit is of the indeterminate form \(\frac{0}{0}\), we simplify the expression using a trigonometric identity

$$ \begin{aligned} \lim_{x\to 0}\frac{\sin 4x}{\sin 2x} = \lim_{x\to 0}\frac{2\sin 2x \cdot \cos 2x}{\sin 2x} \end{aligned} $$

Cancelling the common factor \(\sin 2x\) from the numerator and denominator

$$ \begin{aligned} &= \lim_{x\to 0} 2\cos 2x\\ &= 2\cos 2\cdot 0\\ &= 2\cos 0\\ &= 2\times 1\\ &= 2 \end{aligned} $$

Hence, the value of the given limit is \(2\)

Evaluate: \(\lim\limits_{x\to0}\dfrac{\tan x}{x}\)

Solution

We begin by rewriting the given expression using the identity \(\tan x = \dfrac{\sin x}{\cos x}\)

$$ \begin{aligned} &\lim_{x\to 0}\frac{\tan x}{x}\\ &= \lim_{x\to 0}\frac{\sin x}{x}\cdot \frac{1}{\cos x} \end{aligned} $$

Now, using the standard limit \(\lim_{x\to 0}\dfrac{\sin x}{x} = 1\) and the continuity of the cosine function at \(x=0\)

$$ \begin{aligned} &= \lim_{x\to 0} 1 \cdot \frac{1}{\cos 0}\\ &= \frac{1}{1}\\ &= 1 \end{aligned} $$

Hence, the value of the given limit is \(1\)

Find the derivative of \(\sin x\) at \(x = 0\)

Solution

Let the given function be defined as \(f(x) = \sin x\)

The derivative of \(f(x)\) at a point is defined using the limit

$$ \begin{aligned} f'(x) = \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h} \end{aligned} $$

To find the derivative at \(x=0\), substitute \(x=0\) in the above expression

$$ \begin{aligned} f'(0) &= \lim_{h\to 0}\frac{\sin(0+h)-\sin 0}{h}\\\\ &= \lim_{h\to 0}\frac{\sin h-0}{h} \end{aligned} $$

Using the standard limit \(\lim_{h\to 0}\dfrac{\sin h}{h} = 1\)

$$ \begin{aligned} f'(0) = 1 \end{aligned} $$

Hence, the derivative of \(\sin x\) at \(x=0\) is \(1\)

Find the derivative of \(f(x) = 3\) at \(x = 0\) and at \(x = 3\).

Solution

The given function is a constant function defined by \(f(x)=3\)

To find the derivative at \(x=0\), we use the definition of derivative

$$ \begin{aligned} f'(0) &= \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}\\\\ &= \lim_{h\to 0}\frac{3-3}{h}\\\\ &= \lim_{h\to 0}\frac{0}{h}\\\\ &= 0 \end{aligned} $$

Now, to find the derivative at \(x=3\), again applying the definition of derivative

$$ \begin{aligned} f'(3) &= \lim_{h\to 0}\frac{f(3+h)-f(3)}{h}\\\\ &= \lim_{h\to 0}\frac{3-3}{h}\\\\ &= \lim_{h\to 0}\frac{0}{h}\\\\ &= 0 \end{aligned} $$

Hence, the derivative of the constant function \(f(x)=3\) is zero at both \(x=0\) and \(x=3\)

Find the derivative of \(f(x) = 10x\).

Solution

The given function is defined as \(f(x)=10x\)

The derivative of \(f(x)\) is obtained using the definition of derivative

$$ \begin{aligned} f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{aligned} $$

Substituting \(f(x)=10x\) in the above expression

$$ \begin{aligned} &= \lim_{h\to 0}\frac{10(x+h)-10x}{h}\\\\ &= \lim_{h\to 0}\frac{10x+10h-10x}{h} \end{aligned} $$

Simplifying the expression

$$ \begin{aligned} &= \lim_{h\to 0}\frac{10h}{h}\\\\ &= 10 \end{aligned} $$

Hence, the derivative of the function \(f(x)=10x\) is \(10\)

Find the derivative of \(f(x) = x^2\).

Solution

The given function is defined as \(f(x)=x^{2}\)

Using the definition of derivative

$$ \begin{aligned} f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{aligned} $$

Substituting \(f(x)=x^{2}\) in the above expression

$$ \begin{aligned} &= \lim_{h\to 0}\frac{(x+h)^{2}-x^{2}}{h}\\\\ &= \lim_{h\to 0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h} \end{aligned} $$

Simplifying the expression

$$ \begin{aligned} &= \lim_{h\to 0}\frac{2xh+h^{2}}{h}\\\\ &= \lim_{h\to 0}(2x+h)\\\\ &= 2x \end{aligned} $$

Hence, the derivative of the function \(f(x)=x^{2}\) is \(2x\)

Find the derivative of \(f(x)=\dfrac{1}{x}\)

Solution

The given function is defined as \(f(x)=\dfrac{1}{x}\)

Using the definition of derivative

$$ \begin{aligned} f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{aligned} $$

Substituting \(f(x)=\dfrac{1}{x}\) in the above expression

$$ \begin{aligned} &= \lim_{h\to 0}\frac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\\\\ &= \lim_{h\to 0}\frac{x-(x+h)}{h\,x(x+h)} \end{aligned} $$

Simplifying the numerator

$$ \begin{aligned} &= \lim_{h\to 0}\frac{x-x-h}{h\,x(x+h)}\\\\ &= \lim_{h\to 0}\frac{-h}{h\,x(x+h)} \end{aligned} $$

Cancelling the common factor \(h\) and applying the limit

$$ \begin{aligned} &= \lim_{h\to 0}\frac{-1}{x(x+h)}\\\\ &= -\frac{1}{x^{2}} \end{aligned} $$

Hence, the derivative of the function \(f(x)=\dfrac{1}{x}\) is \(-\dfrac{1}{x^{2}}\)

Find the derivative of \(f(x) = 1 + x + x^2 + x^3 +... + x^{50}\) at \(x = 1\)

Solution

The given function is a polynomial defined by \(f(x)=1+x+x^{2}+\ldots+x^{50}\)

Differentiating term by term using the power rule

$$ \begin{aligned} f'(x) = 0+1+2x+3x^{2}+\ldots+50x^{49} \end{aligned} $$

Now substituting \(x=1\) to find the value of the derivative at \(x=1\)

$$ \begin{aligned} f'(1) = 1+2+3+\ldots+50 \end{aligned} $$

The sum of the first \(50\) natural numbers is given by

$$ \begin{aligned} &= \frac{50\times 51}{2}\\\\ &= 25\times 51\\\\ &= 1275 \end{aligned} $$

Hence, the value of the derivative of the given function at \(x=1\) is \(1275\)

Find the derivative of \(f(x) =\dfrac{x+1}{x}\)

Solution

The given function is defined as \(f(x)=\dfrac{x+1}{x}\)

To find the derivative, we apply the quotient rule which states that for \(f(x)=\dfrac{u}{v}\), the derivative is given by \(\dfrac{v\,u'-u\,v'}{v^{2}}\)

$$ \begin{aligned} f'(x) = \frac{x\cdot \frac{d}{dx}(x+1)-(x+1)\cdot \frac{d}{dx}(x)}{x^{2}} \end{aligned} $$

Differentiating the numerator terms and simplifying

$$ \begin{aligned} &= \frac{x\cdot 1-(x+1)\cdot 1}{x^{2}}\\\\ &= \frac{x-x-1}{x^{2}}\\\\ &= \frac{-1}{x^{2}} \end{aligned} $$

Hence, the derivative of the function \(f(x)=\dfrac{x+1}{x}\) is \(-\dfrac{1}{x^{2}}\)

Compute the derivative of \(\tan x\).

Solution

Let the given function be defined as \(f(x)=\tan x\)

Using the definition of derivative

$$ \begin{aligned} f'(x) = \lim_{h\to 0}\frac{\tan(x+h)-\tan x}{h} \end{aligned} $$

Expressing the tangent function in terms of sine and cosine

$$ \begin{aligned} = \lim_{h\to 0}\frac{1}{h}\left[\frac{\sin(x+h)}{\cos(x+h)}-\frac{\sin x}{\cos x}\right] \end{aligned} $$

Combining the fractions using a common denominator

$$ \begin{aligned} = \lim_{h\to 0}\frac{1}{h}\left[\frac{\cos x\sin(x+h)-\sin x\cos(x+h)}{\cos x\cos(x+h)}\right] \end{aligned} $$

Using the trigonometric identity \(\sin A\cos B-\cos A\sin B=\sin(A-B)\)

$$ \begin{aligned} &= \lim_{h\to 0}\frac{1}{h}\left[\frac{\sin((x+h)-x)}{\cos x\cos(x+h)}\right]\\\\ &= \lim_{h\to 0}\frac{\sin h}{h}\cdot\frac{1}{\cos x\cos(x+h)} \end{aligned} $$

Applying the standard limit \(\lim_{h\to 0}\dfrac{\sin h}{h}=1\) and using the continuity of cosine

$$ \begin{aligned} &= 1\cdot\frac{1}{\cos x\cos x}\\ &= \frac{1}{\cos^{2}x}\\ &= \sec^{2}x \end{aligned} $$

Hence, the derivative of \(\tan x\) is \(\sec^{2}x\)

Compute the derivative of \(f(x) = \sin^2 x\).

Solution

The given function is defined as \(f(x)=\sin^{2}x\)

Writing the function as a product

$$ \begin{aligned} f(x)=\sin x\cdot \sin x \end{aligned} $$

Differentiating using the product rule

$$ \begin{aligned} f'(x) = \sin x\cdot (\sin x)' + (\sin x)'\cdot \sin x \end{aligned} $$

Since \((\sin x)'=\cos x\), we obtain

$$ \begin{aligned} &= \sin x\cos x+\cos x\sin x\\ &= 2\sin x\cos x \end{aligned} $$

Using the trigonometric identity \(2\sin x\cos x=\sin 2x\)

$$ \begin{aligned} f'(x)=\sin 2x \end{aligned} $$

Hence, the derivative of \(f(x)=\sin^{2}x\) is \(\sin 2x\)