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  • <img alt="\small \text { If } f(x)=\left(\frac{x}{2+x}\right)^{2 x} \text {, then } \lim _{x \rightarrow \infty} f(x) \text { is }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20%

\small \text { If } f(x)=\left(\frac{x}{2+x}\right)^{2 x} \text {, then } \lim _{x \rightarrow \infty} f(x) \text { is }

Option: 1

e^{-4}


Option: 2

e^{-6}


Option: 3

e


Option: 4

\frac{1}{e}


Answers (1)

best_answer

\begin{aligned} & f(x)=\left(\frac{x}{x+2}\right)^{2 x} \\ & \log f(x)=\lim _{x \rightarrow \infty} 2 x \log \left(\frac{x+2-2}{x+2}\right)=2 x\left(\log \left(1-\frac{2}{x+2}\right)\right) \\ & =\lim _{x \rightarrow \infty}-1 \times 2 x\left(\frac{2}{x+2}+\frac{1}{2}\left(\frac{2}{x+2}\right)^2+\frac{1}{3}\left(\frac{x}{x+2}\right)^3 \cdots \ldots . .\right) \\ & =\lim _{x \rightarrow \infty}-2 x \frac{2}{x+2}\left(1+\frac{1}{x+2}+\ldots \ldots\right)=\lim _{x \rightarrow \infty} \frac{-4}{1+\frac{2}{x}}=-4 \\ & f(x)=e^{-4} \end{aligned}

Hence (A) is the correct answer.

Posted by

Sayak

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