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\lim_{x\rightarrow 0}\frac{\sin x^{4}-x^{4}\: \cos x^{4}+x^{20}}{x^{4}\left ( e^{2x^{4}}-1-2x^{4} \right )} is equal to

Option: 1

0


Option: 2

-\frac{1}{6}


Option: 3

\frac{1}{6}


Option: 4

does not exist


Answers (1)

best_answer

As we have learnt in

 

Limit of product / quotient -

Limit of product/quotient is the product/quotient of individual limits such that

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

=\lim_{x\rightarrow a}{f(x).\lim_{x\rightarrow a}g(x), given that f(x) and g(x) are non-zero finite values

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}, given that f(x) and g(x) are non-zero finite values

Also\:\lim_{x\rightarrow a}{kf(x)} 

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

 

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\lim_{x\rightarrow 0}\frac{\sin x^{4}-x^{4}\: \cos x^{4}+x^{20}}{x^{4}\left ( e^{2x^{4}}-1-2x^{4} \right )}

=\lim_{t\rightarrow 0}\frac{\sin t-t\: \cos t+t^{5}}{t(e^{2t}-1-2x^{4})} \; \; \left \{ Let\: x^{4}=t \right \}

=\lim_{t\rightarrow 0}\frac{\left ( t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-... \right )-t\left ( 1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-... \right )+t^{5}}{t\left ( 1+2t+\frac{4t^{2}}{2!}+\frac{8t^{3}}{3!}+\frac{16t^{4}}{4!}+...-1-2t \right )}

=\lim_{t\rightarrow 0}\frac{-\frac{t^{3}}{6}+\frac{t^{3}}{2}+\frac{t^{5}}{5!}-\frac{t^{5}}{4!}+...+t^{5}}{2t^{3}+\frac{8t^{4}}{3!}+...}

=\frac{-\frac{1}{6}+\frac{1}{2}}{2}=\frac{-1+3}{12}=\frac{1}{6}

Posted by

Pankaj Sanodiya

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