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Please,please help me - Limit , continuity and differentiability - JEE Main-4

\lim_{x\rightarrow 0} \: \frac{x\tan 2x-2x\tan x}{\left ( 1-\cos 2x \right )^{2}} equals :

  • Option 1)

    \frac{1}{4}

  • Option 2)

    1

  • Option 3)

    \frac{1}{2}

  • Option 4)

    -\frac{1}{2}

 
Answers (2)
117 Views
N neha

As we learned,

 

Limit of product / quotient -

Limit of product is the product of individual limits such that

\lim_{x\rightarrow a}{f(x).g(x)}

=\lim_{x\rightarrow a}{f(x).\lim_{x\rightarrow a}g(x)

also\:\lim_{x\rightarrow a}{kf(x)}

=k\lim_{x\rightarrow a}{f(x)}

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}

 

-

 

 and

 

Evalution of Trigonometric limit -

\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1

\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1

put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0

Then\:it\:comes

\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1

-

 

 

\lim_{x\rightarrow 0} \: \frac{x\tan 2x-2x\tan x}{\left ( 1-\cos 2x \right )^{2}}

\Rightarrow \: \lim_{x\rightarrow 0} \: \frac{x\left ( \frac{2\tan x}{1-\tan ^{2}x}-2\tan x \right )}{ \left ( 2\sin ^{2}x \right ) ^{2}}

 

\Rightarrow \: \lim_{x\rightarrow 0} \: \frac{2tanx\cdot x \left (\frac{1-\left ( 1-\tan ^{2}x \right )}{1-\tan ^{2}x} \right )}{4\sin ^{4}x}

\Rightarrow \: \lim_{x\rightarrow 0} \frac{\frac{2x\tan x\times \tan ^{2}x}{1-\tan ^{2}x}}{4\sin ^{4}x}

\Rightarrow \: \lim_{x\rightarrow 0} \frac{1}{2}\: \frac{x\tan ^{3}x}{\sin x\cdot \sin ^{3}x} \times \frac{1}{1-\tan ^{2}x}

\Rightarrow \: \lim_{x\rightarrow 0} \frac{1}{2}\times 1\times 1=\frac{1}{2}


Option 1)

\frac{1}{4}

Option 2)

1

Option 3)

\frac{1}{2}

Option 4)

-\frac{1}{2}

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