How to compute the limit. My first instinct was to convert the expression in a fraction and use l'hopitals rule, but the didnt seem like it was going anywhere. Are there any better approaches to evaluating this limit?
$$\lim_{x \to \infty}\left[x\left(1+\frac{1}{x}\right)^x-ex\right]$$
Best Answer
$$\ln\left[\left(1+\frac1x\right)^x\right] =x\ln\left(1+\frac1x\right)=1-\frac1{2x}+O(x^{-2})$$ so $$\left(1+\frac1x\right)^x =e\exp\left(-\frac1{2x}+O(x^{-2})\right)=e\left(1-\frac1{2x}+O(x^{-2}) \right)$$ and so $$x\left(1+\frac1x\right)^x-ex\to-\frac e2$$ as $x\to\infty$.