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  • What is the value of <img alt="\lim _{x \rightarrow 0}\left(1+\left(e^x-e^{-x}\right)\right)^{1 / \sin x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%200%7D

What is the value of \lim _{x \rightarrow 0}\left(1+\left(e^x-e^{-x}\right)\right)^{1 / \sin x}?

Option: 1

1


Option: 2

0


Option: 3

e^{2}


Option: 4

e


Answers (1)

best_answer

Given:  \lim _{x \rightarrow 0}\left(1+\left(e^x-e^{-x}\right)\right)^{1 / \sin x}

If \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$, then $e^{\lim _{x \rightarrow a(x)}^{f(x)}}.
=e^{\lim _{x \rightarrow 0} \frac{e^x-e^{-x}}{\sin x}}

By using the expansion formula e^x=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\ldots,the above expression becomes,

=e^{\lim _{x \rightarrow 0} \frac{\left[1+\frac{x}{1 !} +\frac{x^2}{2 !}+\ldots\right]-\left[1-\frac{x}{1 !}+\frac{x^2}{2 !}-\dots \right]}{\sin x}}
=e^{\lim _{x \rightarrow 0} \frac{2\left[\frac{x}{1!}+\frac{x^3}{3 !} +\frac{x^5}{5 !}+\ldots\right]}{\sin x}}

Take out the x as common in the numerator,

=e^{\lim _{x \rightarrow 0} \frac{2x\left[1+\frac{x^{2}}{3!}+\frac{x^{4}}{5 !} +\ldots\right]}{\sin x}}
=e^{\lim _{x \rightarrow 0} \frac{2\left[1+\frac{x^{2}}{3!}+\frac{x^{4}}{5 !} +\ldots\right]}{\frac{\sin x}{x}}}
=e^{\frac{2}{1}}
=e^{2}

Therefore, \lim _{x \rightarrow 0}\left(1+\left(e^x-e^{-x}\right)\right)^{1 / \sin x}= e^{2}.

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Sayak

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