Using several times L'Hospital Rule I got $$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$ Is it possible find this limit without L'Hospital?
2026-04-07 05:54:14.1775541254
How to evaluate $\lim\limits_{x \rightarrow +\infty}{e^x (e - (1+\frac{1}{x} )^x)}$ without L'Hospital?
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The natural thing to do is to look at the logarithm of $\left(1+\frac{1}{x}\right)^x$, that is, at $x\log\left(1+\frac{1}{x}\right)$. Use the series $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots.$$ From this we can obtain good estimates of the difference between $e$ and $(1+1/x)^x$ when $x$ is large. For the calculation, the series expansion of $e^t$ is useful.