Question

Match the limits of the functions given in Column I with their corresponding values given in Column II and choose the correct option from the codes given below.

Column I		Column II
A	$\underset{x\to \pi }{\lim }\frac{\sin (\pi -x)}{\pi (\pi -x)}$	1	4
B	$\underset{x\to 0}{\lim }\frac{\cos x}{\pi -x}$	2	$\frac{1}{\pi }$
C	$\underset{x\to 0}{\lim }\frac{\cos 2x-1}{\cos x-1}$	3	$\frac{a+1}{b}$
D	$\underset{x\to 0}{\lim }\frac{ax+x\cos x}{b\sin x}$	4	0
E	$\underset{x\to 0}{\lim }x\sec x$	5	1
F	$\underset{x\to 0}{\lim }\frac{\sin ax+bx}{ax+\sin bx}$ $(a,b,a+b\ne 0)$

• A. A= 2 ,B= 2 ,C= 1 ,D= 3 ,E= 5 ,F= 4
• B. A= 2 ,B=2 ,C=3 ,D= 1,E= 4 ,F= 5
• C. A= 2 ,B=2 ,C=1 ,D= 4 ,E=3 ,F= 5
• D. A= 2,B= 2,C= 1 ,D=3 ,E=4 ,F= 5

Answer 1

Answer: (C)

A. Given, $\underset{x\to \pi }{\lim }\frac{\sin (\pi -x)}{\pi (\pi -x)}$
Let $\pi -x=h$, As $x\to \pi$, then $h\to 0$
$\therefore \underset{x\to \pi }{\lim }\frac{\sin (\pi -x)}{\pi (\pi -x)}=\underset{h\to 0}{\lim }\frac{\sin h}{\pi h}=\underset{h\to 0}{\lim }\frac{1}{\pi }\times \frac{\sin h}{h}$
$=\frac{1}{\pi }\times 1=\frac{1}{\pi }\left(\because \underset{h\to 0}{\lim }\frac{\sin h}{h}=1\right)$
B. Given $\underset{x\to 0}{\lim }\frac{\cos x}{\pi -x}$
Put the limit directly, we get $\frac{\cos 0}{\pi -0}=\frac{1}{\pi }$
C. Given, $\underset{x\to 0}{\lim }\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{\lim }\frac{1-\cos 2x}{1-\cos x}$
$=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{2{\sin }^2\frac{x}{2}}$
$\left(\because 1-\cos 2x=2{\sin }^{2}x\text{ and }1-\cos x=2{\sin }^2\frac{x}{2}\right)$
Multiplying and dividing by $x^2$ and then multiplying by $\frac{4}{4}$ in the numerater. we get
$=\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{x^2}\times \frac{4\times \frac{x^2}{4}}{{\sin }^2\frac{x}{2}}$
$=\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2\times {\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)}^2\times 4$
$=1\times 1\times 4=4$
D. Given, $\underset{x\to 0}{\lim }\frac{ax+x\cos x}{b\sin x}$
Dividing each term by $x$, we get
$-\underset{x\to 0}{\lim }\frac{\frac{ax}{x}+\frac{x\cos x}{x}}{\frac{b\sin x}{x}}$
$=\underset{x\to 0}{\lim }\frac{a+\cos x}{b\left(\frac{\sin x}{x}\right)}$
$=\frac{a+\cos 0}{b\times 1}=\frac{a+1}{b}\left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$
Note In most of the times, when we put the limit, it becomes an indeterminate form. Firstly, simplity the function by expanding or by using the formula such that we get the factor which is causing the function to indeterminate form and then remove the cause.
E. $\underset{x\to 0}{\lim }x\sec x=0\times \sec 0=0\times 1=0$
F. $\underset{x\to 0}{\lim }\frac{\sin ax+bx}{ax+\sin bx}$
Dividing each term by $x$, we get
$=\underset{x\to 0}{\lim }\frac{\frac{\sin ax}{x}+\frac{bx}{x}}{\frac{ax}{x}+\frac{\sin bx}{x}}=\underset{x\to 0}{\lim }\frac{\frac{a\sin ax}{ax}+b}{a+\frac{b\sin bx}{bx}}$
$=\frac{a\times 1+b}{a+b\times 1}=\frac{a+b}{a+b}=1\left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$