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Limits Continuity and Differentiability question

2016 · 10 Apr · Shift 1 · Q42

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JEE Main
Mathematics
Limits Continuity and Differentiability
2016 · 10 Apr · Shift 1 · Q42

Limits Continuity and Differentiability question

2016 · 10 Apr · Shift 1 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}

is :

A
$-$ 2
B
$-{1 \over 2}$
C
${1 \over 2}$
D
2

View written solutionFree

Correct answer: A

We need to evaluate $L=\lim_{x\to 0}\frac{(1-\cos 2x)^2}{2x\tan x-x\tan 2x}.$
First simplify the numerator using the identity $1-\cos 2x=2\sin^2 x.$ So, $(1-\cos 2x)^2=(2\sin^2 x)^2=4\sin^4 x.$ Hence,

Simplify the denominator: $2x\tan x-x\tan 2x=x(2\tan x-\tan 2x).$ Using $\tan 2x=\frac{2\tan x}{1-\tan^2 x},$ we get

Take common factor $2\tan x$ :

=2\tan x\left(\frac{1-\tan^2 x-1}{1-\tan^2 x}\right) =2\tan x\left(\frac{-\tan^2 x}{1-\tan^2 x}\right).

Thus,

So denominator becomes

Substitute into the limit:

=\lim_{x\to 0}-2\sin^4 x\cdot \frac{1-\tan^2 x}{x\tan^3 x}.

Now use $\tan x=\frac{\sin x}{\cos x}.$ Then $\tan^3 x=\frac{\sin^3 x}{\cos^3 x},$ so

Hence,

Now apply standard limits:

Therefore,

So the correct option is $\boxed{\text{A: }-2}.$
Comparison with stored answer: Stored correct answer is A, which matches our result.

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