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  • Solve     <img alt="\mathrm{\lim _{x \rightarrow \infty}\left(x-x^2 \ln \frac{1+x}{x}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Cri

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

Option: 1

\frac{1}{2}


Option: 2

0


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

Let
\mathrm{\lim _{x \rightarrow \infty}\left\{x-x^2 \ln \left(\frac{1+x}{x}\right)\right\} }\mathrm{\lim _{x \rightarrow \infty}\left\{x-x^2 \ln \left(\frac{1+x}{x}\right)\right\} }
Now substitute  \mathrm{x=\frac{1}{h}}
\mathrm{\lim _{h \rightarrow 0}\left\{\frac{1}{h}-\frac{1}{h^2} \ln (1+h)\right\}
Now using Taylor series expansion of   \mathrm{\ln (1+x)}
\mathrm{\lim _{h \rightarrow 0}\left\{\frac{1}{h}-\frac{1}{h^2}\left(h-\frac{h^2}{2}+\frac{h^3}{3}-\cdots\right)\right\} }

Now first 2 terms will get cancelled and after applying limit,
\mathrm{\lim _{h \rightarrow 0}\left\{\frac{1}{h}-\frac{1}{h}+\frac{1}{2}-\frac{h}{3}+\frac{h^2}{4}-\cdots\right\} }
=\frac{1}{2}.

Posted by

avinash.dongre

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