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  • Evaluating    <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5(x)\right)}{\left(e^{x^4}-1\right)\left(\sin \left(x^2\right)-x^2\right)}}" src="https://entrancecor

Evaluating    \mathrm{\lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5(x)\right)}{\left(e^{x^4}-1\right)\left(\sin \left(x^2\right)-x^2\right)}}

Option: 1

0


Option: 2

1


Option: 3

-2


Option: 4

-3


Answers (1)

best_answer

\mathrm{\lim _{t \rightarrow 0} \frac{1-\cos t}{t^2}=\frac{1}{2}, \quad \lim _{t \rightarrow 0} \frac{e^t-1}{t}=1, \quad \lim _{t \rightarrow 0} \frac{\sin t-t}{t^3}=-\frac{1}{6} }

From the first you get
\mathrm{\frac{1}{2}=\lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5 x\right)}{\sin ^{10} x}=\lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5 x\right)}{x^{10}} \frac{x^{10}}{\sin ^{10} x} }

And so

\mathrm{ \lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5 x\right)}{x^{10}}=\frac{1}{2} }

From the second you get
\mathrm{ \lim _{x \rightarrow 0} \frac{e^{x^4}-1}{x^4}=1 }

and from the third that
\mathrm{ \lim _{x \rightarrow 0} \frac{\sin \left(x^2\right)-x^2}{x^6}=-\frac{1}{6} }

Now rewrite your limit as
\mathrm{ \lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5(x)\right)}{\left(e^{x^4}-1\right)\left(\sin \left(x^2\right)-x^2\right)}=\lim _{x \rightarrow 0} \frac{1-\cos \left(\sin ^5 x\right)}{x^{10}} \frac{x^4}{e^{x^4}-1} \frac{x^6}{\sin \left(x^2\right)-x^2}=\frac{1}{2} \cdot 1 \cdot(-6) }\mathrm{ =-3}

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