Question

Find the limit of [1+(2/x^2)]^2x as x approaches infinity

Answer 1

Compare this to the limit that yields e

Answer 2

\[\lim_{x\to\infty}\left(1+\frac{2}{x^2}\right)^{2x}\\
\large e^{\ln\left[\lim_{x\to\infty}\left(1+\frac{2}{x^2}\right)^{2x}\right]}\\
\large e^{\lim_{x\to\infty}\ln\left(1+\frac{2}{x^2}\right)^{2x}}\\
\large e^{\lim_{x\to\infty}2x\ln\left(1+\frac{2}{x^2}\right)}\]
Focusing on just the exponent for a moment:
\[\lim_{x\to\infty}2x\ln\left(1+\frac{2}{x^2}\right)\\
2\lim_{x\to\infty}\frac{\ln\left(1+\frac{2}{x^2}\right)}{\frac{1}{x}}=\frac{0}{0}\]
Applying L'Hopital's rule, you get
\[2\huge \lim_{x\to\infty}\frac{ \frac{-\frac{4}{x^3}}{1+\frac{2}{x^2}} }{-\frac{1}{x^2}}\\
2\large \lim_{x\to\infty}\frac{ \frac{4}{x} }{ 1+\frac{2}{x^2}}=\frac{0}{1}=0\\
\]So, you have
\[\large e^{\displaystyle \lim_{x\to\infty}\cdots}=e^0=1\]