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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?

$$\lim_{x \to \infty}\left[x\left(1+\frac{1}{x}\right)^x-ex\right]$$

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$$\ln\left[\left(1+\frac1x\right)^x\right] =x\ln\left(1+\frac1x\right)=1-\frac1{2x}+O(x^{-2})$$ 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 $x\to\infty$.

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  • $\begingroup$ What happened here? $e\exp\left(-\frac1{2x}+O(x^{-2})\right)=e\left(1-\frac1{2x}+O(x^{-2}) \right)$ $\endgroup$ – Anirudh Jul 9 at 12:38
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L'Hôpital's rule is fine, just write $$\lim_{x\to \infty} \frac{\left(1+\frac{1}{x}\right)^x-e}{\frac{1}{x}}$$

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By letting $t=1/x\to 0$, we have that $$\begin{align} \lim_{x \to \infty} \left((1+\frac{1}{x})^xx-ex\right)&=\lim_{x \to \infty} \frac{\exp(x\log(1+\frac{1}{x}))-e}{1/x}\\ &=e\lim_{t \to 0} \frac{\exp\left(\frac{\log(1+t)}{t}-1\right)-1}{t}\\ &=e\lim_{t \to 0} \frac{\exp\left(\frac{\log(1+t)}{t}-1\right)-1}{\frac{\log(1+t)}{t}-1}\cdot \frac{\frac{\log(1+t)}{t}-1}{t}\\ &=e\cdot 1\cdot\lim_{t \to 0} \frac{\log(1+t)-t}{t^2}=-\frac{e}{2} \end{align}.$$

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Let $x=1/t$, the $$L=\lim_{t\rightarrow 0}\frac{(1+t)^{1/t}-e}{t.}$$ The Mclaurin expansion of $$(1+t)^{1/t}=e-et/2+11et^2/24+....$$ Then $$L=\lim_{t\rightarrow 0}\frac{-et/2+11et/24}{t}=-e/2$$

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Writing $y=1/x$ you want:

$$\lim_{y\to 0^+}\frac{(1+y)^{1/y}-e}{y}.$$

Now you can apply L'Hopital. Since $f(y)=(1+y)^{1/y}=\exp\left(\frac{\log(1+y)}{y}\right)$ the derivative of the numerator is $$f'(y)=\left(\frac{1}{(1+y)y}-\frac{\log(1+y)}{y^2}\right)f(y)=\frac{y-(y+1)\log(y+1)}{(y+1)y^2}f(y)$$

We know $f(y)\to e$ and $\frac{1}{1+y}\to 1$ as $y\to 0^+.$ So:

$$\lim_{y\to 0^+}f'(y)=e\lim_{y\to 0^+}\frac{y-(y+1)\log(y+1)}{y^2}$$

Now, $\log(1+y)=y-\frac{y^2}{2}+O(y^3).$ So $$\begin{align}y-(y+1)\log(1+y)&=y-y\log(y+1)-\log(y+1)\\&=y-\left(y^2+O(y^3)\right)-\left(y-y^2/2+O(y^3)\right)\\&=-y^2/2+O(y^3).\end{align}$$

So $$\frac{y-(y+1)\log(y+1)}{y^2}=-\frac{1}{2}+O(y).$$

So the limit is $-\frac{e}{2}.$

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