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Evaluate \mathrm{\lim _{x \rightarrow \infty} \frac{x^2 \cdot\left(1+\frac{1}{\ln ^2 x}\right)^{1+2 \ln ^2 x}-x^2+1}{x-2+x^2}}

Option: 1

\mathrm{ e^2-1 }


Option: 2

\mathrm{ e^2+1 }


Option: 3

0


Option: 4

\mathrm{ 2 e^2+1 }


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow \infty} \frac{x^2\left(1+\frac{1}{(\ln (x))^2}\right)^{\left(1+2(\ln (x))^2\right)}-x^2+1}{x^2+x-2}}
\mathrm{=\lim _{x \rightarrow \infty} \frac{\left(1+\frac{1}{(\ln (x))^2}\right)^{\left(1+2(\ln (x))^2\right)}-1+\frac{1}{x^2}}{1+\frac{1}{x}-\frac{2}{x^2}}}
Then we have
\mathrm{\lim _{x \rightarrow \infty}\left(1+\frac{1}{(\ln (x))^2}\right)^{\left(1+2(\ln (x))^2\right)}}
\mathrm{=\exp \left(\lim _{x \rightarrow \infty} \frac{1+2(\ln (x))^2}{(\ln (x))^2}\right)}
\mathrm{=\exp \left(\lim _{x \rightarrow \infty} \frac{1}{(\ln (x))^2}+2\right)}
\mathrm{ =\exp (2)=e^2 }
So finally we have 
\mathrm{ \lim _{x \rightarrow \infty} \frac{x^2\left(1+\frac{1}{(\ln (x))^2}\right)^{\left(1+2(\ln (x))^2\right)}-x^2+1}{x^2+x-2}=e^2-1 }

 

Posted by

Pankaj

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