A) 1
B) \[-\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
Given limit is \[L=\underset{x\to 0}{\mathop{\lim }}\,\frac{(x\tan 2x-2x\tan x)}{{{(1-\cos 2x)}^{2}}}\] By expanding \[\tan 2x\] and \[\cos 2x\] we get \[\frac{(x\tan 2x-2x\tan x)}{{{(1-\cos 2x)}^{2}}}=\frac{x\frac{2\tan x}{1-{{(\tan x)}^{2}}}-2x\tan x}{{{(1-(1-2{{\sin }^{2}}x))}^{2}}}\] \[=\frac{2x\tan x-[2x\tan x-2x{{\tan }^{3}}x]}{4{{\sin }^{4}}x\times (1-{{\tan }^{2}}x)}=\frac{2x{{\tan }^{3}}x}{4{{\sin }^{4}}x\times (1-{{\tan }^{2}}x)}\] \[=\frac{2x{{\tan }^{3}}x}{4{{\sin }^{4}}x\times (\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x})}=\frac{2x\frac{{{\sin }^{3}}x}{{{\cos }^{3}}x}}{4{{\sin }^{4}}x\times (\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x})}\] \[=\frac{x}{2\sin x\times ({{\cos }^{2}}x-{{\sin }^{2}}x)\cos x}\] Now applying the limit \[x\to 0\], we get \[L=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{2\sin x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos x({{\cos }^{2}}x-{{\sin }^{2}}x)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{2\sin x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\cos 0({{\cos }^{2}}0-{{\sin }^{2}}0)}\] \[=\frac{1}{2}[\because \underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}=1]\] Hence, option D is correct. Let us have \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}}=l\] \[l=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( \frac{2\tan x}{1-{{\tan }^{2}}x} \right)-2x\tan x}{{{\left( 1-\frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)}^{2}}}\] \[l=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x\tan x\left( \frac{1}{1-{{\tan }^{2}}x}-1 \right)}{{{\left( \frac{2{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)}^{2}}}\] \[l=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x\tan x\left( \frac{{{\tan }^{2}}x}{1-{{\tan }^{2}}x} \right)}{{{\left( \frac{2{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)}^{2}}}\] \[l=\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{(1+{{\tan }^{2}}x)}^{2}}}{2\tan x(1-{{\tan }^{2}}x)}\] \[l=\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{({{\sec }^{2}}x)}^{2}}\cos x}{2\sin x(1-{{\tan }^{2}}x)}\] We know \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}=1\] \[\therefore l=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
