Math Assignment Class XI Ch-13

Math Assignment Class XI | Chapter 13 | Limits

Math Assignment Class XI Chapter 13 Limits and derivatives, Extra questions on Limits strictly according to the CBSE & DAV Board, Important questions on limits.

MATHS ASSIGNMENT ON LIMIT

Question 1 :

Evaluate the limits: $\displaystyle \lim_{x \to 2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}$

Ans: -1/4

Question 2 :

Evaluate the limits: $\displaystyle \lim_{x \to 0}\frac{\sqrt{1-x}-1}{x}$

Ans: - 1/2

Question 3:

Evaluate the limits $\lim_{x\rightarrow 0}\frac{cot2x-cosec2x}{x}$

Ans: -1

Question 4 :

Evaluate the limits $\lim_{x\rightarrow 0}\frac{sinx-2sin3x+sin5x}{x}$

Ans: 0

Question 5:

Evaluate the limits $\lim_{x\rightarrow 0}\; \frac{sin2x+sin6x}{sin5x-sin3x}$

Ans: 4

Question 6 :

Evaluate the limits $\lim_{x\rightarrow 0}\;\; \frac{sin^{2}3x}{x^{2}}$

Ans: 9

Question 7 :

If f(x) = $\left\{\begin{matrix}3x+2, x\leq 0 \\\\\; \; 2(x+1), x> 0\end{matrix}\right.$ then find $\displaystyle \lim_{x \to 0}f(x)$

Ans: 2

Solution Hint: Find LHL and RHL

Question 8:

Evaluate the limits: $\displaystyle \lim_{x \to 0}\frac{Log(1+3x)}{x}$

Ans: 3

Solution Hint:

Use formula: $\displaystyle \lim_{x \to 0}\frac{Log(1+x)}{x}=1$

Question 9 :

If $\displaystyle \lim_{ x\to 0}\frac{sin3x}{sin7x} = K$ , then find the value of K

Ans: 3/7

Question 10 :

Evaluate the limits: $\lim_{x\rightarrow 9 }\;\; \frac{x^{3/2}-27}{x-9}$

Ans: 9/2

Question 11 :

If $\displaystyle \lim_{x \to 1}\frac{x^{4}-1}{x-1}=\displaystyle \lim_{x \to k}\frac{x^{3}-k^{3}}{x^{2}-k^{2}}$ then find the value of k

Ans: 8/3

Question 12 :

Evaluate the limits: $\lim_{x\rightarrow 4 }\;\; \frac{x^{2}-16}{\sqrt{x^{2}+9}-5}$

Ans: 10

Question 13 :

Evaluate the limits: $\lim_{x\rightarrow 2 }\;\; \frac{x^{2}-4}{\sqrt{3x-2}-\sqrt{x+2}}$

Ans: 8

Question 14 :

Evaluate the limits: $\lim_{x\rightarrow 4 }\;\; \frac{(x^{2}-x-12)^{18}}{(x^{3}-8x^{2}+16x)^{9}}$

Ans: $Ans:\left ( \frac{7}{2} \right )^{18}$

Question 15 :

Evaluate the limits: $\lim_{x\rightarrow 1 }\;\; \left [ \frac{2}{1-x^{2}}+\frac{1}{x-1} \right ]$

Ans: 1/2

Question 16 :

Evaluate the limits: $\lim_{x\rightarrow 2 }\;\; \frac{x^{3}-6x^{2}+11x-6}{x^{2}-6x+8}$

Ans: 1/2

Question 17 :

Evaluate the limits: $\lim_{x\rightarrow 2 }\;\; \frac{x^{3}-3x^{2}+4}{x^{4}-8x^{2}+16}$

Ans: 3/16

Question 18 :

Evaluate the limits $\lim_{x\rightarrow 1/2 }\;\; \frac{8x^{3}-1}{16x^{4}-1}$

Ans: 3/4

Question 19 :

Evaluate the limits $\lim_{x\rightarrow 0}\left [ \frac{\sqrt{1+tanx}-\sqrt{1-tanx}}{sin2x} \right ]$

Ans: 1/2

Solution Hint:

Rationalizing the numerator.

Simplify and then putting the limit.

Question 20 $\lim_{x\rightarrow \frac{\pi }{3}}\frac{\sqrt{3}-tanx}{\pi -3x}$

Ans: 4/3

Question 21

Evaluate the limits: $\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sinx}{cos^{2}x}$

Ans: 1/2

Question 22

Evaluate the limits: $\lim_{x\rightarrow \pi }\frac{sin3x-3sinx}{(\pi -x)^{3}}$

Ans: - 4

Solution Hint

Use Sin3x = 3sinx – 4sin³x,

putting x = π+h, as x→ π, then h→0

Question 23

Evaluate the limits: $\displaystyle \lim_{x \to 0}\frac{1-cos2x}{x(e^{5x}-1)}$

Ans: 2/5

Solution Hint: $\displaystyle \lim_{x \to 0}\frac{1-cos2x}{x(e^{5x}-1)}$ = $\displaystyle \lim_{x \to 0}\frac{2sin^{2}x}{5x(\frac{e^{5x}-1}{5})}$

= $\displaystyle \lim_{x \to 0}\frac{\frac{2sin^{2}x}{x^{2}}}{5(\frac{e^{5x}-1}{5x})}$ = 2/5

Question 24

Evaluate the limits: $\displaystyle \lim_{x \to 3}\frac{x^{2}-5x+6}{x-3}$

Ans: 1

Question 25

Evaluate the limits: $\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{cosx-sinx}{x-\frac{\pi }{4}}$

Ans: - √2

Solution Hint:

$\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{cosx-sinx}{x-\frac{\pi }{4}}$ = $\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{\sqrt{2}\left ( \frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx \right )}{x-\frac{\pi }{4}}$

= $\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{\sqrt{2}\left (sin \frac{\pi }{4}cosx-cos\frac{\pi }{4}sinx \right )}{x-\frac{\pi }{4}}$

= $\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{\sqrt{2}sin\left ( \frac{\pi }{4}-x \right )}{x-\frac{\pi }{4}}$

= $\displaystyle \lim_{x \to \frac{\pi }{4}}\frac{-\sqrt{2}sin\left (x- \frac{\pi }{4} \right )}{x-\frac{\pi }{4}}$

= $-\sqrt{2}\times 1$ = $-\sqrt{2}$

Question 26

Evaluate the limits: $\displaystyle \lim_{x \to \frac{\pi }{6}}\left [ \frac{\sqrt{3}sinx-cosx}{x-\frac{\pi }{6}} \right ]$

Ans: 2

Solution Hint:

= $\displaystyle \lim_{x \to \frac{\pi }{6}}\left [ \frac{2\left ( \frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx \right )}{x-\frac{\pi }{6}} \right ]$

= $\displaystyle \lim_{x \to \frac{\pi }{6}}\left [ \frac{2\left ( cos\frac{{\pi }}{6}sinx-sin\frac{\pi }{6}cosx \right )}{x-\frac{\pi }{6}} \right ]$

= $\displaystyle \lim_{x \to \frac{\pi }{6}}2\left [ \frac{sin\left (x-\frac{{\pi }}{6} \right )}{x-\frac{\pi }{6}} \right ]$

=2 x 1 = 2

Question 27

Evaluate the limits: $\displaystyle \lim_{x \to 0}\frac{10^{x}-5^{x}-2^{x}+1}{x^{2}}$

Ans: Log5.Log2

Solution Hint:

= $\displaystyle \lim_{x \to 0}\frac{2^{x}\times 5^{x}-5^{x}-2^{x}+1}{x^{2}}$

= $\displaystyle \lim_{x \to 0}\frac{5^{x}( 2^{x}-1)-1(2^{x}-1)}{x^{2}}$

= $\displaystyle \lim_{x \to 0}\frac{(5^{x}-1)( 2^{x}-1)}{x^{2}}$

= $\displaystyle \lim_{x \to 0}\left ( \frac{5^{x}-1}{x} \right )\displaystyle \lim_{x \to 0}\left ( \frac{2^{x}-1}{x} \right )$

= Log5.Log2

Formula used : $\displaystyle \lim_{x \to 0}\frac{e^{x}-1}{x}=Loge=1$

Question 28

Evaluate the limits: $\displaystyle \lim_{x \to 2}\frac{cos2x-cos3x}{cos4x-1}$

Ans: -5/16

Solution Hint:

= $\displaystyle \lim_{x \to 2}\frac{-(cos3x-cos2x)}{-(1-cos4x)}$

= $\displaystyle \lim_{x \to 0}\frac{-2sin\frac{5x}{2}sin\frac{x}{2}}{2sin^{2}2x}$

= $-\displaystyle \lim_{x \to 0}\frac{sin\frac{5x}{2}}{x}\times \frac{sin\frac{x}{2}}{x}\div \left ( \frac{sin2x}{x} \right )^{2}$

= $-\displaystyle \lim_{x \to 0}\left ( \frac{sin\frac{5x}{2}}{\frac{5x}{2}}\times \frac{5}{2} \right ) \left ( \frac{sin\frac{x}{2}}{\frac{x}{2}}\times \frac{1}{2} \right )\div \left ( \frac{sin2x}{2x} \times 2\right )^{2}$