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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)
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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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