Q1. \(\lim\limits_{x\to3} x+3\)

Solution

We evaluate the limit by directly substituting the value of \(x\) because the expression \(x+3\) is a polynomial and hence is continuous for all real values of \(x\).

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

Since the function is continuous at \(x=3\), the limit is equal to the value of the function at that point.

Q2. \(\lim\limits_{x\to\pi}\left(x-\dfrac{22}{7}\right)\)

Solution

The given expression is a linear function of \(x\), which is defined and continuous for all real values of \(x\). Therefore, the limit can be evaluated by direct substitution of \(x=\pi\).

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

Hence, the value of the limit is obtained by replacing \(x\) with \(\pi\), as there is no discontinuity at that point.

Q3. \(\lim\limits_{r\to1}\pi r^2\)

Solution

The given expression \(\pi r^2\) is a polynomial function of \(r\) multiplied by a constant. Such functions are continuous for all real values of \(r\). Hence, the limit can be evaluated by direct substitution of \(r=1\).

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

Therefore, since the function is continuous at \(r=1\), the value of the limit is equal to the value of the function at that point.

Q4. \(\lim\limits_{x\to4}\dfrac{4x+3}{x-2}\)

Solution

The given rational function is defined at \(x=4\) because the denominator does not become zero at this value. Therefore, the limit can be evaluated by direct substitution of \(x=4\).

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

Since the function is continuous at \(x=4\), the value of the limit is equal to the value of the function at that point.

Q5. \(\lim\limits_{x\to-1}\dfrac{x^{10}+x^5+1}{x-1}\)

Solution

The given expression is a rational function. At \(x=-1\), the denominator \(x-1\) does not become zero, so the function is defined and continuous at this point. Hence, the limit can be evaluated by direct substitution.

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

Therefore, since there is no discontinuity at \(x=-1\), the value of the limit is equal to the value of the function at that point.

Q6. \(\lim\limits_{x\to0}\dfrac{(x+1)^5-1}{x}\)

Solution

On direct substitution of \(x=0\), the numerator and denominator both become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, further simplification is required.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{(x+1)^5-1}{x} \end{aligned} \]

Expanding \((x+1)^5\) using the binomial theorem, we get

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

Dividing each term in the numerator by \(x\), the expression becomes

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

Now the limit can be evaluated by direct substitution since the expression is a polynomial.

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

Therefore, the value of the given limit is \(5\).

Q7. \(\lim\limits_{x\to2}\dfrac{3x^2-x-10}{x^2-4}\)

Solution

On direct substitution of \(x=2\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, algebraic simplification is necessary to evaluate the limit.

\[ \begin{aligned} \lim_{x\rightarrow 2}\dfrac{3x^{2}-x-10}{x^{2}-4} &= \dfrac{3\cdot 2^{2}-2-10}{4-4} \\ &= \dfrac{12-12}{0} \\ &= \dfrac{0}{0} \end{aligned} \]

To remove the indeterminate form, we factor both the numerator and the denominator.

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

Now the expression is continuous at \(x=2\), so the limit can be found by direct substitution.

\[ \begin{aligned} \dfrac{3\cdot 2+5}{2+2} = \dfrac{11}{4} \end{aligned} \]

Therefore, the value of the given limit is \(\dfrac{11}{4}\).

Q8. \(\lim\limits_{x\to3}\dfrac{x^4-81}{2x^2-5x-3}\)

Solution

On direct substitution of \(x=3\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, factorization is required to evaluate the limit.

\[ \begin{aligned} &\lim_{x\rightarrow 3}\dfrac{x^{4}-81}{2x^{2}-5x-3} \\ &= \dfrac{3^{4}-81}{2\cdot 3^{2}-5\cdot 3-3} \\ &= \dfrac{81-81}{18-15-3} \\ &= \dfrac{0}{0} \end{aligned} \]

We now factorize the numerator and the denominator to remove the indeterminate form.

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

\[ \begin{aligned} 2x^{2}-5x-3 &= 2x^{2}-6x+x-3 \\ &= 2x(x-3)+1(x-3) \\ &= (x-3)(2x+1) \end{aligned} \]

Substituting these factorizations into the given expression and cancelling the common factor \((x-3)\), we get

\[ \begin{aligned} &\lim_{x\rightarrow 3}\dfrac{(x^{2}+9)(x+3)(x-3)}{(x-3)(2x+1)} \\ &= \lim_{x\rightarrow 3}\dfrac{(x^{2}+9)(x+3)}{2x+1} \end{aligned} \]

Now the expression is continuous at \(x=3\), so we substitute \(x=3\) directly.

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

Therefore, the value of the given limit is \(\dfrac{108}{7}\).

Q9. \(\lim\limits_{x\to0}\dfrac{ax+b}{cx+1}\)

Solution

The given rational function is defined at \(x=0\) because the denominator does not become zero at this value. Since the function is continuous at \(x=0\), the limit can be evaluated by direct substitution.

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

Hence, the value of the given limit is b.

Q10. \(\lim\limits_{z\to1}\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\)

Solution

On direct substitution of \(z=1\), the numerator and denominator both become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, a suitable substitution is used to simplify the expression.

\[ \begin{aligned} \lim_{z\rightarrow 1}\dfrac{z^{\tfrac{1}{3}}-1}{z^{\tfrac{1}{6}}-1} \end{aligned} \]

Let \(u=z^{\tfrac{1}{6}}\). Then \(z^{\tfrac{1}{3}}=u^{2}\). Substituting in the given expression, we obtain

\[ \begin{aligned} \dfrac{u^{2}-1}{u-1} &= \dfrac{(u-1)(u+1)}{u-1} \\ &= u+1 \end{aligned} \]

As \(z\rightarrow 1\), we have \(u=z^{\tfrac{1}{6}}\rightarrow 1\). Therefore,

\[ \begin{aligned} \lim_{z\rightarrow 1}\left(z^{\tfrac{1}{6}}+1\right) &= 1+1 \\ &= 2 \end{aligned} \]

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

Q11. \(\lim\limits_{x\to1}\dfrac{ax^2+bx+c}{cx^2+bx+a},\; a+b+c\ne0\)

Solution

The given expression is a rational function. Since \(a+b+c \ne 0\), the denominator does not become zero at \(x=1\). Hence, the function is defined and continuous at this point, allowing direct substitution.

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

Therefore, under the given condition, the value of the limit is \(1\).

Q12. \(\lim\limits_{x\to-2}\dfrac{\frac{1}{x}+\frac{1}{2}}{x+2}\)

Solution

On direct substitution of \(x=-2\), the numerator and the denominator both become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, algebraic simplification is required to evaluate the limit.

\[ \begin{aligned} \lim_{x\rightarrow -2}\dfrac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2} \\ = \dfrac{0}{0} \end{aligned} \]

We simplify the expression by combining the terms in the numerator.

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

Cancelling the common factor \((x+2)\) from the numerator and denominator, we obtain

\[ \begin{aligned} \lim_{x\rightarrow -2}\dfrac{1}{2x} \end{aligned} \]

Now the expression is continuous at \(x=-2\), so we substitute directly.

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

Therefore, the value of the given limit is \(-\dfrac{1}{4}\).

Q13. \(\lim\limits_{x\to0}\dfrac{\sin ax}{bx}\)

Solution

On direct substitution of \(x=0\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, we simplify the expression using a standard trigonometric limit.

\[ \begin{aligned} &\lim_{x\rightarrow 0}\dfrac{\sin ax}{bx} \\ &= \dfrac{0}{0} \end{aligned} \]

We rewrite the expression by multiplying and dividing suitably so that the standard limit \(\lim_{t\to 0}\dfrac{\sin t}{t}=1\) can be applied.

\[ \begin{aligned} &\lim_{x\rightarrow 0}\dfrac{\sin ax}{bx} \\ &= \lim_{x\rightarrow 0}\left(\dfrac{\sin ax}{ax}\cdot\dfrac{a}{b}\right) \end{aligned} \]

As \(x\rightarrow 0\), we have \(ax\rightarrow 0\), so \(\dfrac{\sin ax}{ax}\rightarrow 1\).

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{\sin ax}{bx} &= 1\cdot\dfrac{a}{b} \\ &= \dfrac{a}{b} \end{aligned} \]

Therefore, the value of the given limit is \(\dfrac{a}{b}\).

Q14. \(\lim\limits_{x\to0}\dfrac{\sin ax}{\sin bx},\;a, b \ne0\)

Solution

On direct substitution of \(x=0\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, we simplify the expression using standard trigonometric limits.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{\sin ax}{\sin bx} \end{aligned} \]

We rewrite the expression by multiplying and dividing appropriately so that the standard limit \(\lim_{t\to 0}\dfrac{\sin t}{t}=1\) can be applied to both numerator and denominator.

\[ \begin{aligned} &\lim_{x\rightarrow 0}\dfrac{\sin ax}{\sin bx} \\ &= \lim_{x\rightarrow 0}\left(\dfrac{\sin ax}{ax}\cdot\dfrac{ax}{bx}\cdot\dfrac{bx}{\sin bx}\right) \end{aligned} \]

As \(x\rightarrow 0\), we have \(ax\rightarrow 0\) and \(bx\rightarrow 0\), so \(\dfrac{\sin ax}{ax}\rightarrow 1\) and \(\dfrac{bx}{\sin bx}\rightarrow 1\).

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{\sin ax}{\sin bx} &= 1\cdot\dfrac{a}{b}\cdot 1 \\ &= \dfrac{a}{b} \end{aligned} \]

Therefore, for \(a\ne 0\) and \(b\ne 0\), the value of the given limit is \(\dfrac{a}{b}\).

Q15. \(\lim\limits_{x\to\pi}\dfrac{\sin (\pi-x)}{\pi(\pi-x)}\)

Solution

On direct substitution of \(x=\pi\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, we simplify the expression using a standard trigonometric identity.

\[ \begin{aligned} \lim_{x\rightarrow \pi}\dfrac{\sin(\pi-x)}{\pi(\pi-x)} \end{aligned} \]

Using the identity \(\sin(\pi-x)=\sin x\), the expression becomes

\[ \begin{aligned} &\lim_{x\rightarrow \pi}\dfrac{\sin(\pi-x)}{\pi(\pi-x)} \\ &= \dfrac{1}{\pi}\lim_{x\rightarrow \pi}\dfrac{\sin(\pi-x)}{\pi-x} \end{aligned} \]

Let \(t=\pi-x\). Then as \(x\rightarrow \pi\), we have \(t\rightarrow 0\). Substituting, we get

\[ \begin{aligned} \dfrac{1}{\pi}\lim_{t\rightarrow 0}\dfrac{\sin t}{t} \end{aligned} \]

Using the standard limit \(\lim_{t\to 0}\dfrac{\sin t}{t}=1\), we obtain

\[ \begin{aligned} &\lim_{x\rightarrow \pi}\dfrac{\sin(\pi-x)}{\pi(\pi-x)} \\ &= \dfrac{1}{\pi} \end{aligned} \]

Therefore, the value of the given limit is \(\dfrac{1}{\pi}\).

Q16. \(\lim\limits_{x\to0}\dfrac{\cos x}{\pi-x}\)

Solution

The given function is defined at \(x=0\) because the denominator \(\pi-x\) does not become zero at this value. Also, \(\cos x\) is continuous for all real values of \(x\). Hence, the limit can be evaluated by direct substitution.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{\cos x}{\pi-x} &= \dfrac{\cos 0}{\pi-0} \\ &= \dfrac{1}{\pi} \end{aligned} \]

Therefore, the value of the given limit is \(\dfrac{1}{\pi}\).

Q17. \(\lim\limits_{x\to0}\dfrac{\cos 2x-1}{\cos x-1}\)

Solution

On direct substitution of \(x=0\), both the numerator and the denominator become zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, algebraic simplification using trigonometric identities is required.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{\cos 2x-1}{\cos x-1} \end{aligned} \]

We multiply and divide the expression by \((\cos x+1)\) to simplify the denominator.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{(\cos 2x-1)(\cos x+1)}{(\cos x-1)(\cos x+1)} \end{aligned} \]

Using the identity \(\cos^2 x-1=\tfrac{1}{2}(\cos 2x-1)\times 2\), the denominator becomes \(\cos^2 x-1\), and we rewrite the expression as

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{2(\cos 2x-1)(\cos x+1)}{2(\cos^2 x-1)} \end{aligned} \]

Since \(\cos^2 x-1=\tfrac{1}{2}(\cos 2x-1)\), we cancel the common factor \((\cos 2x-1)\).

\[ \begin{aligned} \lim_{x\rightarrow 0}2(\cos x+1) \end{aligned} \]

Now the expression is continuous at \(x=0\), so we substitute directly.

\[ \begin{aligned} 2(\cos 0+1) &= 2(1+1) \\ &= 4 \end{aligned} \]

Therefore, the value of the given limit is \(4\).

Q18. \(\lim\limits_{x\to0}\dfrac{ax+x\cos x}{b\sin x}\)

Solution

On direct substitution of \(x=0\), both the numerator and the denominator tend to zero, so the given limit is of the indeterminate form \(\tfrac{0}{0}\). Hence, we simplify the expression using standard trigonometric limits.

\[ \begin{aligned} \lim_{x\rightarrow 0}\dfrac{ax+x\cos x}{b\sin x}\cdot\dfrac{x}{x} \end{aligned} \]

Factoring \(x\) from the numerator and rewriting, we obtain

\[ \begin{aligned} &\lim_{x\rightarrow 0}\dfrac{x(a+\cos x)}{b\sin x}\cdot\dfrac{x}{x} \\ &= \lim_{x\rightarrow 0}\dfrac{a+\cos x}{b\left(\dfrac{\sin x}{x}\right)} \end{aligned} \]

As \(x\rightarrow 0\), we have \(\cos x\rightarrow 1\) and \(\dfrac{\sin x}{x}\rightarrow 1\).

\[ \begin{aligned} &\lim_{x\rightarrow 0}\dfrac{a+\cos x}{b} \\ &= \dfrac{a+1}{b} \end{aligned} \]

Therefore, the value of the first limit is \(\dfrac{a+1}{b}\).

Q19. \(\lim\limits_{x\to0}x\sec x\)

Solution

We evaluate \(\lim_{x\to0} x\sec x\) by direct substitution.

Since \(\sec x = \frac{1}{\cos x}\), rewrite the expression as \(\frac{x}{\cos x}\).

As \(x \to 0\), the numerator approaches 0 while the denominator approaches \(\cos 0 = 1\).

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

The limit exists and equals 0 because both the function and its limit value are continuous at \(x = 0\).

Q20. \(\lim\limits_{x\to0}\dfrac{\sin ax+bx}{ax+\sin bx},\; a,b,a+b\ne0\)

Solution

We evaluate \(\lim_{x\to0}\frac{\sin ax+bx}{ax+\sin bx}\) using standard sine limits, given \(a,b,a+b\neq0\).

Divide numerator and denominator by \(x\) to get \(\frac{\frac{\sin ax}{x}+b}{a+\frac{\sin bx}{x}}\).

Now rewrite using known limits \(\lim_{x\to0}\frac{\sin ax}{ax}=1\) and \(\lim_{x\to0}\frac{\sin bx}{bx}=1\), so multiply and divide appropriately by \(ax\) and \(bx\).

$$\begin{aligned} \lim_{x\to 0}&\frac{\sin ax+bx}{ax+\sin bx}\\ &=\lim_{x\to 0}\frac{\left( \frac{\sin ax}{ax}+\frac{bx}{ax}\right) }{\left( \frac{ax}{bx}+\frac{\sin bx}{bx}\right) }\frac{ax}{bx}\\ &=\frac{\left( 1+\frac{b}{a}\right) }{\left( \frac{a}{b}+1\right) }\times \frac{a}{b}\\ &=\frac{\left( a+b\right) b}{a\left( a+b\right) }\cdot \frac{a}{b}\\ &=1 \end{aligned}$$

The steps leverage the continuity of the standard limits, confirming the value simplifies to 1 independently of specific \(a\) and \(b\) values under the given conditions.

Q21. \(\lim\limits_{x\to0}(\text{cosec }x-\cot x)\)

Solution

We evaluate \(\lim_{x\to0}(\text{cosec } x - \cot x)\) by simplifying the expression first.

Rewrite using definitions: \(\text{cosec } x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\), yielding \(\frac{1 - \cos x}{\sin x}\).

This is an indeterminate form \(\frac{0}{0}\); rationalize numerator using \(1 - \cos x = 2\sin^2\frac{x}{2}\) and \(\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}\), but here we multiply cleverly to simplify.

$$\begin{aligned} \lim_{x\to 0} &\text{cosec } x - \cot x\\ &=\lim_{x\to 0}\frac{1 - \cos x}{\sin x}\\ &=\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\\ &=\lim_{x\to 0}\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\\ &=\frac{0}{1}\\ &=0 \end{aligned}$$

The algebraic simplification resolves the indeterminate form correctly, confirming the limit is 0 as \(x\to0\).

Q22. \(\lim\limits_{x\to\frac{\pi}{2}}\dfrac{\tan 2x}{x-\frac{\pi}{2}}\)

Solution

We evaluate \(\lim_{x\to\frac{\pi}{2}}\frac{\tan 2x}{x-\frac{\pi}{2}}\) by substitution to simplify the argument.

Let \(y = x - \frac{\pi}{2}\), so as \(x\to\frac{\pi}{2}\), \(y\to0\) and \(x = y + \frac{\pi}{2}\).

Substitute to get \(\lim_{y\to0}\frac{\tan 2\left(y + \frac{\pi}{2}\right)}{y} = \lim_{y\to0}\frac{\tan\left(2y + \pi\right)}{y}\).

$$\begin{aligned} \lim_{x\to\frac{\pi}{2}} &\frac{\tan 2x}{x-\frac{\pi}{2}}\\ y &= x-\frac{\pi}{2}\\ x &= y+\frac{\pi}{2}\\ \lim_{y\to 0} &\frac{\tan 2\left(y+\frac{\pi}{2}\right)}{y}\\ &= \lim_{y\to 0}\frac{\tan(2y + \pi)}{y}\\ \tan(\theta + \pi) &= \tan \theta\\ \lim_{y\to 0} &\frac{\tan 2y}{y}\\ &= \lim_{y\to 0} 2\cdot\frac{\tan 2y}{2y}\\ &= 2\cdot 1\\ &= 2 \end{aligned}$$

The key periodicity property \(\tan(\theta + \pi) = \tan \theta\) reduces it to the standard limit \(\lim_{\theta\to0}\frac{\tan \theta}{\theta} = 1\), yielding 2.

Q23. Find \(\lim\limits_{x\to0}f(x)\) and \(\lim\limits_{x\to1}f(x)\), where \( \begin{aligned} f(x)=\begin{cases} 2x+3, \;x\le0\\ 3(x+1), \;x\gt0 \end{cases} \end{aligned} \)

Solution

For the limit as x → 0, the definition of the function changes at x = 0, so the left-hand and right-hand limits are examined separately. When x ≤ 0, the expression for f(x) is 2x + 3, and when x > 0, the expression is 3(x + 1). Since both one-sided limits give the same finite value, the limit at x = 0 exists.

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

Hence, \(\lim\limits_{x\to0} f(x) = 3\).

For the limit as x → 1, all values of x sufficiently close to 1 satisfy x > 0, so only the second expression of the function is relevant.

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

Therefore, \(\lim\limits_{x\to1} f(x) = 6\).

Q24. Find \(\lim\limits_{x\to1}f(x)\), where \( \begin{aligned} f(x)=\begin{cases} x^2-1, \;x\le0\\ -x^2-1, \;x\gt0 \end{cases} \end{aligned} \)

Solution

Since the point x = 1 lies entirely in the region x > 0, the same expression of the function applies for values of x approaching 1 from both the left and the right. Hence, both one-sided limits must be evaluated using f(x) = -x² − 1.

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

Since the left-hand and right-hand limits are equal, the limit exists and \(\lim\limits_{x\to1} f(x) = −2\).

Q25. Evaluate \(\lim\limits_{x\to0}f(x)\) where \( \begin{aligned} f(x)=\begin{cases} \dfrac{\mid x \mid}{x}, \;x\ne0\\ 0, \;x=0 \end{cases} \end{aligned} \)

Solution

To evaluate the limit as x → 0, the left-hand and right-hand limits are considered separately because the expression |x| / x depends on the sign of x.

$$ \begin{aligned} \lim_{x \to 0^-} \dfrac{|x|}{x} &= \dfrac{-x}{x} \\ &= -1 \\\\ \lim_{x \to 0^+} \dfrac{|x|}{x} &= \dfrac{x}{x} \\ &= 1 \\\\ f(0) &= 0 \\ \lim_{x \to 0^-} &\ne \lim_{x \to 0^+} \ne f(0) \end{aligned} $$

Since the left-hand limit and right-hand limit are not equal, the limit of f(x) as x → 0 does not exist, even though the function is defined at x = 0.

Q26. Find \(\lim\limits_{x\to0}f(x)\) where \( \begin{aligned} f(x)=\begin{cases} \dfrac{ x }{\mid x \mid}, \;x\ne0\\ 0, \;x=0 \end{cases} \end{aligned} \)

Solution

To find the limit as x → 0, the left-hand and right-hand limits are evaluated separately because the value of x / |x| depends on whether x is negative or positive.

$$ \begin{aligned} \lim_{x \to 0^-} \dfrac{x}{|x|} &= \dfrac{x}{-x} \\ &= -1 \\\\ \lim_{x \to 0^+} \dfrac{x}{|x|} &= \dfrac{x}{x} \\ &= 1 \\\\ f(0) = 0 \\ \lim_{x \to 0^-} \ne \lim_{x \to 0^+} \ne f(0) \end{aligned} $$

Since the left-hand limit and right-hand limit are not equal, the limit of f(x) as x → 0 does not exist, even though the function is defined at x = 0.

Q27. Find \(\lim\limits_{x\to5}f(x)\) where \(f(x)=\mid x \mid-5\)

Solution

The function f(x) = |x| − 5 is continuous at x = 5. Therefore, the limit of the function as x approaches 5 can be found by direct substitution.

$$ \begin{aligned} \lim_{x \to 5} |x| - 5 &= |5| - 5 \\ &= 5 - 5 \\ &= 0 \end{aligned} $$

Hence, \(\lim\limits_{x\to5}f(x) = 0\).

Q28. Suppose \( \begin{aligned} f(x)=\begin{cases} a+bx, & x\le0\\ 4, & x=1\\ b-ax,& x\gt1 \end{cases} \end{aligned} \)

and if \(\lim\limits_{x\to1}f(x)=f(1)\) what are possible values of a and b?

Solution

Since \(\lim\limits_{x\to1} f(x) = f(1)\), the function must be continuous at x = 1. Hence, the left-hand limit, the right-hand limit, and the value of the function at x = 1 must all be equal.

$$ \begin{aligned} \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} (a + bx) \\ &= a + b \\\\ \lim_{x \to 1^+} f(x) &= \lim_{x \to 1^+} (b - ax) \\ &= b - a \\\\ f(1) &= 4 \\ a + b &= 4 \\ b - a &= 4 \\ 2b &= 8 \\ b &= 4 \\ a + b &= 4 \\ a &= 4 - b \\ &= 4 - 4 \\ &= 0 \\ a &= 0 \\ b &= 4 \end{aligned} $$

Therefore, the possible values of the constants are a = 0 and b = 4, for which the function is continuous at x = 1.

Q29. Let \(a_1 ,\; a_2 ,\; . . .,\; a_n\) be fixed real numbers and define a function

\(f(x)=(x-a_1)(x-a_2)\ldots(x-a_n)\),

What is \(\lim\limits_{x\to a_1}f(x)\)? for some \(a\ne a_1,\;a_2,\,\ldots\; a_n\), compute \(\lim\limits_{x\to a}f(x)\)

Solution

The function f(x) is a polynomial, and hence it is continuous for all real values of x. Therefore, the limit at any point can be evaluated by direct substitution.

$$ \begin{aligned} &\lim_{x \to a_1} (x-a_1)(x-a_2)\ldots(x-a_n) \\ &= (a_1-a_1)(a_1-a_2)\ldots(a_1-a_n) \\ &= 0 \cdot (a_1-a_2)\ldots(a_1-a_n) \\ &= 0 \end{aligned} $$

Now let a be any real number different from a_1, a_2, \ldots, a_n. Since none of the factors becomes zero at x = a, the limit is obtained by direct substitution.

$$ \begin{aligned} &\lim_{x \to a} (x-a_1)(x-a_2)\ldots(x-a_n) \\ &= (a-a_1)(a-a_2)\ldots(a-a_n) \end{aligned} $$

Hence, the limit is zero at x = a_1, and for any other value a, the limit equals the value of the polynomial at x = a.

Q30. If \( \begin{aligned} f(x)=\begin{cases} \mid x\mid + 1, & x\lt0\\ 0, & x=0\\ \mid x\mid-1,& x\gt0 \end{cases} \end{aligned} \)

For what value (s) of a does \(\lim\limits_{x\to a}f(x)\) exists?

Solution

To determine the values of a for which \(\lim\limits_{x\to a f(x)\) exists, we examine the behavior of the function in each region defined by its piecewise form.

$$ \begin{aligned} f(x) &= |x| + 1 \quad \text{for } x < 0 \\ f(x) &=0 \quad \text{for } x=0 \\ f(x) &=|x| - 1 \quad \text{for } x> 0 \end{aligned} $$

For any value of a such that a < 0, the function is given by |x| + 1, which is continuous. Hence, the limit exists for all a < 0. Similarly, for any a > 0, the function is given by |x| − 1, which is also continuous, so the limit exists for all a > 0.

The only point requiring special attention is a = 0, where the definition of the function changes. The left-hand and right-hand limits are evaluated separately.

$$ \begin{aligned} \lim_{x \to 0^-} f(x) &= \lim_{x \to 0^-} (|x| + 1) \\ &= 1 \\\\ \lim_{x \to 0^+} f(x) &= \lim_{x \to 0^+} (|x| - 1) \\ &= -1 \end{aligned} $$

Since the left-hand limit and the right-hand limit at x = 0 are not equal, the limit does not exist at this point.

Therefore, \(\lim\limits_{x\to a} f(x)\) exists for all real values of a except a = 0.

Q31. If the function \(f(x)\) satisfies \(\lim\limits_{x\to1}\dfrac{f(x)-2}{x^2-1}=\pi\), evaluate \(\lim\limits_{x\to0}f(x)\) and \(\lim\limits_{x\to1}f(x)\) exists?

Solution

We are given that

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

As x → 1, the denominator x² − 1 tends to zero. Since the given limit is a finite real number, the numerator f(x) − 2 must also tend to zero as x → 1. Hence,

$$ \begin{aligned} \lim_{x \to 1} (f(x) - 2) &= 0 \\ \lim_{x \to 1} f(x) &= 2 \end{aligned} $$

Therefore, the limit of f(x) as x → 1 exists and is equal to 2.

Now consider \(\lim\limits_{x\to 0} f(x)\). The given information relates the behavior of f(x) only in a neighborhood of x = 1. No condition is provided about the behavior of the function near x = 0.

Hence, the limit \(\lim\limits_{x\to 0} f(x)\) cannot be determined from the given information.

Thus, \(\lim\limits_{x\to1} f(x)\) exists and equals 2, while \(\lim\limits_{x\to0} f(x)\) cannot be evaluated from the given data.

Q32. If \( \begin{aligned} f(x)=\begin{cases} mx^2+n, & x\lt0\\ nx+m, & 0\le x \le 1\\ nx^3+m,& x\gt1 \end{cases} \end{aligned} \), For what integers \(m\) and \(n\) does both \(\lim\limits_{x\to0}f(x)\) and \(\lim\limits_{x\to1}f(x)\) exists?

Solution

For the existence of the limits at x = 0 and x = 1, the left-hand and right-hand limits at each point must be equal.

First, consider the limit as x → 0. The expressions on either side of 0 are different, so the one-sided limits are evaluated separately.

$$ \begin{aligned} \lim_{x \to 0^-} f(x) &= \lim_{x \to 0^-} (mx^2 + n) \\ &= n \\\\ \lim_{x \to 0^+} f(x) &= \lim_{x \to 0^+} (nx + m) \\ &= m \end{aligned} $$

For the limit at x = 0 to exist, these must be equal, giving

$$ \begin{aligned} n = m \end{aligned} $$

Next, consider the limit as x → 1. The function is defined by \(nx + m\) for x ≤ 1 and by \(nx^3 + m\) for x > 1.

$$ \begin{aligned} \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} (nx + m) \\ &= n + m \\\\ \lim_{x \to 1^+} f(x) &= \lim_{x \to 1^+} (nx^3 + m) \\ &= n + m \end{aligned} $$

Since both one-sided limits are equal for all real values of n and m, the limit at x = 1 exists automatically.

Combining both conditions, the limits at x = 0 and x = 1 exist when

$$ \begin{aligned} m = n \end{aligned} $$

Hence, for all integers m and n satisfying m = n, both required limits exist.