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  • Evaluate    <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}}{x^3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightar

Evaluate    \mathrm{\lim _{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}}{x^3}}

Option: 1

0


Option: 2

\frac{1}{6}


Option: 3

\frac{2}{3}


Option: 4

\frac{1}{4}


Answers (1)

best_answer

If the limit exists, let it be L, by the change of variable \mathrm{x \rightarrow-x}  and taking the average,
\mathrm{ L=\lim _{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}}{x^3}=\lim _{x \rightarrow 0} \frac{e^{-x}-1+x-\frac{x^2}{2}}{-x^3}=\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3} . }

Now with \mathrm{ x \rightarrow 3 x. }, and using  \mathrm{ \frac{\sinh x}{x} \rightarrow 1^* }


\mathrm{ L=\lim _{x \rightarrow 0} \frac{\sinh 3 x-3 x}{27 x^3}=\lim _{x \rightarrow 0} \frac{3 \sinh x+4 \sinh ^3 x-3 x}{27 x^3}=\frac{1}{9} L+\frac{4}{27} . }
Hence,

\mathrm{ L=\frac{1}{6} \text {. } }
 

Posted by

vishal kumar

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