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  • Find the limit of the following: <img alt="\lim _{x \rightarrow 0}(1+x \cos x-\log (1+x))^{1 / x^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%200%7D%281a

Find the limit of the following: \lim _{x \rightarrow 0}(1+x \cos x-\log (1+x))^{1 / x^2}

Option: 1

1


Option: 2

0


Option: 3

\frac{2}{n}


Option: 4

\sqrt{e}


Answers (1)

best_answer

Given: \lim _{x \rightarrow 0}(1+x \cos x-\log (1+x))^{1 / x^2}

If  \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$, then $e^{\lim _{x \rightarrow a} \frac{f(x)}{g(x)}}.
= e^{\lim_{x\rightarrow 0}\left ( \frac{x\cos x-\log\left ( 1+x \right )}{x^{2}} \right )}

Applying L-Hospital’s rule, the equation becomes,

=e^{\lim _{x \rightarrow 0}\left(\frac{\cos x-x \sin x-1 /(1+x)}{2 x}\right)}
=e^{\lim _{x \rightarrow 0}\left(\frac{-\sin x-\sin x-x \cos x-1 /(1+x)^2}{2}\right)}
=e^{1 / 2}
=\sqrt{e}

Therefore,\lim _{x \rightarrow 0}(1+x \cos x-\log (1+x))^{1 / x^2}=\sqrt{e}.

 

Posted by

Rakesh

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