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

Last Updated : 07 Apr, 2021
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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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