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  • Give answer! - is equal to : - Limit , continuity and differentiability - JEE Main

\lim_{x\rightarrow 0} \frac{x \cot (4x)}{\sin ^{2}x \cot ^{2}(2x)}  is equal to :

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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

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\lim_{x\rightarrow 0}\frac{x\cot4x }{\sin ^{2}x\cot ^{2}2x}=\lim_{x\rightarrow 0}\frac{x\tan^{2}2x }{\sin ^{2}x\tan 4x}

                                      =\lim_{x\rightarrow 0}\frac{x(\frac{\tan^{2}2x}{4x^{2}}) \times 4x^{2}}{\frac{\sin ^{2}x}{x^{2}}\times x^{2}(\frac{\tan 4x}{4x} )\times 4x}

                                     =1\left ( \lim_{z\rightarrow 0} \right )\frac{\tan z}{z}=1\: \: \: \: \: \left ( \lim_{z\rightarrow 0} \right )\frac{\sin z}{z}=0

 


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