How to find solution of UP PGT mathematics previous years question

Question : Find solution of UP PGT mathematics previous years questionSolution :To evaluate \[limit_{x\rightarrow\infty}\left(\frac{x}{1+x}\right)^{2x}=\left(limit_{x\rightarrow\infty}\left(\frac{1+x-1}{1+x}\right)^x\right)^2\] We can write it as \[=\left(limit_{1+x\rightarrow\infty}\left(1-\frac{1}{1+x}\right)^{1+x}\right)^2 \]and this becomes \[=\left(e^{-1}\right)^2\] which is equal to \[=e^{-2}\] Therefore \[limit_{x\rightarrow\infty}\left(\frac{x}{1+x}\right)^{2x}=\frac{1}{e^2}\] Hence correct Option is (4).

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