Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 1

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Evaluate the following limits in Exercises 1 to 22.

Question 1: $\lim_{x \to 3} x+3$

Solution:

In $\lim_{x \to 3} x+3$ , as x⇢3
Put x = 3, we get
$\lim_{x \to 3} x+3$ = 3+3
= 6

Question 2: $\lim_{x \to \pi} (x-\frac{22}{7})$

Solution:

In $\lim_{x \to \pi} (x-\frac{22}{7})$ , as x⇢π
Put x = π, we get
$\lim_{x \to \pi} (x-\frac{22}{7}) = (π-\frac{22}{7})$
= $(π-\frac{22}{7})$

Question 3: $\lim_{r \to 1} \pi r^2$

Solution:

In $\lim_{r \to 1} \pi r^2$ , as r⇢1
Put r = 1, we get
$\lim_{r \to 1} \pi r^2 = \pi (1)^2$
= π

Question 4: $\lim_{x \to 4} (\frac{4x+3}{x-2})$

Solution:

In $\lim_{x \to 4} (\frac{4x+3}{x-2})$ , as x⇢4
Put x = 4, we get
$\lim_{x \to 4} (\frac{4x+3}{x-2}) = \frac{4(4)+3}{4-2}$
= $\frac{19}{2}$

Question 5: $\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})$

Solution:

In $\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})$ , as x⇢-1
Put x = -1, we get
$\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1}) = \frac{(-1)^{10}+(-1)^5+1}{-1-1}$
= $\frac{1-1+1}{-2}$
= $\frac{-1}{2}$

Question 6: $\lim_{x \to 0} \frac{(x+1)^5-1}{x}$

Solution:

In $\lim_{x \to 0} \frac{(x+1)^5-1}{x}$ , as x⇢0
Put x = 0, we get
$\lim_{x \to 0} \frac{(x+1)^5-1}{x} = \frac{(0+1)^5-1}{0} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s take x+1=p and x = p-1, to make it equivalent to theorem.
$\mathbf{\lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1}}$
As, x⇢0 ⇒ p⇢1
$\lim_{p \to 1} \frac{(p)^5-1}{p-1} = \lim_{p \to 1} \frac{(p)^5-1^5}{p-1}$
Here, n=5 and a = 1.
$\lim_{p \to 1} \frac{(p)^5-1}{p-1} = 5(1)^{5-1}$
= 5(1)⁴
= 5

Question 7: $\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$

Solution:

In $\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$ , as x⇢2
Put x = 2, we get
$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4} = \frac{3(2)^2-2-10}{2^2-4} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s Factorise the numerator and denominator, we get
$\lim_{x \to 2} \frac{3x^2-6x+5x-10}{x^2-4}$
= $\lim_{x \to 2} \frac{(3x+5)(x-2)}{(x+2)(x-2)}$
Cancelling (x-2), we have
= $\lim_{x \to 2} \frac{(3x+5)}{(x+2)}$
Put x = 2, we get
$\lim_{x \to 2} \frac{(3x+5)}{(x+2)} = \frac{(3(2)+5)}{(2+2)}$
= $\frac{11}{4}$

Question 8: $\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}$

Solution:

In $\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}$ , as x⇢3
Put x = 3, we get
$\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3} = \frac{(3)^4-81}{2(3)^2-5(3)-3} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s Factorise the numerator and denominator, we get
$\lim_{x \to 3} \frac{(x^2)^2-9^2}{2x^2-6x+x-3}$
= $\lim_{x \to 3} \frac{(x^2-9)(x^2+9)}{(2x+1)(x-3)}$
= $\lim_{x \to 3} \frac{(x+3)(x-3)(x^2+9)}{(2x+1)(x-3)}$
Cancelling (x-3), we have
= $\lim_{x \to 3} \frac{(x+3)(x^2+9)}{(2x+1)}$
Put x = 3, we get
= $\lim_{x \to 3} \frac{(x+3)(x^2+9)}{(2x+1)} = \frac{(3+3)(3^2+9)}{(2(3)+1)}$
= $\frac{9\times 18}{7}$
= $\frac{108}{7}$

Question 9: $\lim_{x \to 0} \frac{ax+b}{cx+1}$

Solution:

In $\lim_{x \to 0} \frac{ax+b}{cx+1}$ , as x⇢0
Put x = 0, we get
$\lim_{x \to 0} \frac{ax+b}{cx+1} = \frac{a(0)+b}{c(0)+1}$
= b

Question 10: $\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Solution:

In $\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$ , as z⇢1
Put z = 1, we get
$\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1} = \frac{1^{\frac{1}{3}}-1}{1^{\frac{1}{6}}-1} = \frac{0}{0}$
Let’s take $z^{\frac{1}{6}}$ = p and $z^{\frac{1}{3}}$ = p²,
As, z⇢1 ⇒ p⇢1
= $\lim_{p \to 1} \frac{p^2-1}{p-1}$
Now, let’s Factorise the numerator, we get
= $\lim_{p \to 1} \frac{(p-1)(p+1)}{p-1}$
Cancelling (p-1), we have
= $\lim_{p \to 1} (p+1)$
Put p = 1, we get
$\lim_{p \to 1} (p+1) = 1+1$
= 2

Question 11: $\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a},\hspace{0.1cm}a+b+c\neq0$

Solution:

In $\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a}$ , as x⇢1
Put x = 1, we get
$\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a} = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a}$
= $\frac{a+b+c}{c+b+a}$
= 1 (As it is given a+b+c≠0)

Question 12: $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}$

Solution:

In $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}$ , as x⇢-2
Firstly, lets simplify the equation
$\frac{\frac{1}{x} + \frac{1}{2}}{x+2} = \frac{\frac{2+x}{2x}}{x+2}$
Cancelling (x+2),we get
$\lim_{x \to -2} \frac{1}{2x}$
Put x = -2, we get
$\lim_{x \to -2} \frac{1}{2x} = \frac{1}{2(-2)}$
= $\frac{-1}{4}$

Question 13: $\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}$

Solution:

In $\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}$ , as x⇢0
Put x = 0, we get
$\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx} = \frac{sin\hspace{0.1cm}a(0)}{b(0)} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s multiply and divide the equation by a, to make it equivalent to theorem.
$\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}$
Hence, we have
$\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx} \times \frac{a}{a}$
= $\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{ax} \times \frac{a}{b}$
= $\frac{a}{b} \lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{ax}$
As x⇢0, then ax⇢0
= $\frac{a}{b} \lim_{ax \to 0} \frac{sin\hspace{0.1cm}ax}{ax}$
By using the theorem, we get
= $\frac{a}{b} . 1$
= $\frac{a}{b}$

Question 14: $\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx},\hspace{0.1cm}a,b\neq0$

Solution:

In $\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx}$ , as x⇢0
Put x = 0, we get
$\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx} = \frac{sin \hspace{0.1cm}a(0)}{sin \hspace{0.1cm}b(0)} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s multiply and divide the numerator by ax and denominator by bx to make it equivalent to theorem.
$\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}$
Hence, we have
$\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}(ax) \times ax}{ax}}{\frac{sin \hspace{0.1cm}(bx) \times bx}{bx}}$
= $\frac{a}{b}\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}ax}{ax}}{\frac{sin \hspace{0.1cm}bx}{bx}}$
= $\frac{a}{b} \frac{\lim_{x \to 0}\frac{sin \hspace{0.1cm}ax}{ax}}{\lim_{x \to 0}\frac{sin \hspace{0.1cm}bx}{bx}}$
By using the theorem, we get
= $\frac{a}{b} . 1 .1$
= $\frac{a}{b}$

Question 15: $\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}$

Solution:

In $\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}$ , as x⇢π
Put x = π, we get
$\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)} = \frac{sin(\pi-\pi)}{\pi(\pi-\pi)} = \frac{0}{0}$
As, this limit becomes undefined
Now, let’s take π-x=p
As, x⇢π ⇒ p⇢0
$\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}$
= $\lim_{p \to 0} \frac{sin p}{p\pi}$
= $\frac{1}{\pi} \lim_{p \to 0} \frac{sin\hspace{0.1cm} p}{p}$
By using the theorem, we get
= $1. \frac{1}{\pi}$
= $\frac{1}{\pi}$

Question 16: $\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}$

Solution:

In $\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}$ , as x⇢0
Put x = 0, we get
$\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)} = \frac{cos\hspace{0.1cm}0}{(\pi-0)}$
= $\frac{1}{\pi}$

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Last Updated : 07 Apr, 2021

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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 1

Evaluate the following limits in Exercises 1 to 22.

Question 1: $\lim_{x \to 3} x+3$

Question 2: $\lim_{x \to \pi} (x-\frac{22}{7})$

Question 3: $\lim_{r \to 1} \pi r^2$

Question 4: $\lim_{x \to 4} (\frac{4x+3}{x-2})$

Question 5: $\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})$

Question 6: $\lim_{x \to 0} \frac{(x+1)^5-1}{x}$

Question 7: $\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$

Question 8: $\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}$

Question 9: $\lim_{x \to 0} \frac{ax+b}{cx+1}$

Question 10: $\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Question 11: $\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a},\hspace{0.1cm}a+b+c\neq0$

Question 12: $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}$

Question 13: $\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}$

Question 14: $\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx},\hspace{0.1cm}a,b\neq0$

Question 15: $\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}$

Question 16: $\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 1

Evaluate the following limits in Exercises 1 to 22.

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:

Question 16:

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 1: $\lim_{x \to 3} x+3$

Question 2: $\lim_{x \to \pi} (x-\frac{22}{7})$

Question 3: $\lim_{r \to 1} \pi r^2$

Question 4: $\lim_{x \to 4} (\frac{4x+3}{x-2})$

Question 5: $\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})$

Question 6: $\lim_{x \to 0} \frac{(x+1)^5-1}{x}$

Question 7: $\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$

Question 8: $\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}$

Question 9: $\lim_{x \to 0} \frac{ax+b}{cx+1}$

Question 10: $\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Question 11: $\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a},\hspace{0.1cm}a+b+c\neq0$

Question 12: $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}$

Question 13: $\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}$

Question 14: $\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx},\hspace{0.1cm}a,b\neq0$

Question 15: $\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}$

Question 16: $\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}$