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  • Evaluating    <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{x^2 \cos x-6 \ln \left(1+x^2\right)+5 x^2}{\left(e \sqrt[4]{1+4 x^3+8 x^4}-e\right) \arcsin (x)}}" src="https://entrance

Evaluating    \mathrm{\lim _{x \rightarrow 0} \frac{x^2 \cos x-6 \ln \left(1+x^2\right)+5 x^2}{\left(e \sqrt[4]{1+4 x^3+8 x^4}-e\right) \arcsin (x)}}

Option: 1

\mathrm{5 / 2 e}


Option: 2

\mathrm{2e}


Option: 3

\mathrm{\frac{e}{2}}


Option: 4

0


Answers (1)

best_answer

We write the nuemreator as \mathrm{x^2(\cos x-1)-6\left(\ln \left(1+x^2\right)-x^2\right)} then dividing top and bottom by \mathrm{x^4} the limit becomes

                                           \mathrm{ \lim _{x \rightarrow 0} \frac{\frac{\cos x-1}{x^2}-\frac{6}{x^4}\left(\ln \left(1+x^2\right)-x^2\right)}{\frac{\arcsin x}{x} \cdot \frac{e^{\sqrt{1+x^3+8 x^4-1}-1}}{x^3}} }

The numerator can be computed using the limits \mathrm{\lim _{x \rightarrow 0} \frac{\cos x-1}{x^2}=-\frac{1}{2}} and \mathrm{\lim _{x \rightarrow 0} \frac{\ln \left(1+x^2\right)-x^2}{x^4}=\frac{-1}{2}}where the

latter is proved in Are all limits solvable without L'Hôpital Rule or Series Expansion

For the bottom, we use the limits in the original post except now it is more divided by \mathrm{x^3}

\mathrm{ \begin{gathered} \lim _{x \rightarrow 0} \frac{e^{\sqrt[4]{1+4 x^3+8 x^4}}-e}{x^3}=e \lim _{x \rightarrow 0} \frac{e^{\sqrt[4]{1+4 x^3+8 x^4}-1}-1}{\sqrt[4]{1+4 x^3+8 x^4}-1} \cdot \frac{\sqrt[4]{1+4 x^3+8 x^4}-1}{\left(1+4 x^3+8 x^4\right)-1} \cdot \frac{4 x^3+8 x^4}{x^3}=e \\ \cdot \frac{1}{4} \cdot 4 \end{gathered} }
Combining all of this we arrive at the desired result \mathrm{\frac{5}{2 e}}

Posted by

Irshad Anwar

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