Finding: $\displaystyle\lim_{x\to\infty}x\big((1+\tfrac1x)^x-\mathrm{e}\big) $

Question

Find: $$\displaystyle\lim_{x\to\infty}x\Big(\big(1+\tfrac1x\big)^x-\mathrm{e}\Big) $$

EDIT: so we have here a $\infty\cdot0$ so I'll try LHR, Edit2: I don't think LHR will get me anywhere since it will always be zero in the denominator...

Maybe develop a Taylor series ?

Wrong try:

$\displaystyle\lim_{x\to\infty}(1+\frac1x)^x=\frac 1e$ is a well known limit so:

$\lim_{x\to\infty}x(\displaystyle\lim_{x\to\infty}(1+\frac1x)^x-e)=\lim_{x\to\infty}x(\frac 1e-e)=-\infty $

Answer 1

Note that $$\lim_{x\to\infty}\left(1+\frac 1x\right)^x=e.$$

You can use l'Hôpital's rule.

Yours will be

$$\begin{align}\lim_{x\to\infty}\frac{(1+(1/x))^x-e}{(1/x)}&=\lim_{x\to\infty}\frac{(1+(1/x))^x \{-1+(1+x) \ln(1+(1/x))\}}{-(x+1)/x^2}\\&=\lim_{x\to\infty}\frac{(1+(1/x))^x \{-x+x(1+x) \ln(1+(1/x))\}}{-(x+1)/x}\\&=\frac{e\times (1/2)}{-1+0}\\&=-e/2\end{align}$$ where $$\begin{align}\lim_{x\to\infty}\{-x+x(1+x)\ln(1+1/x)\}&=\lim_{x\to\infty}\frac{-1+(1+x)\ln(1+(1/x))}{1/x}\\&=\lim_{x\to\infty}\frac{\ln(1+(1/x))-(1/x)}{-1/x^2}\\&=\lim_{x\to\infty}\frac{1/(x^2+x^3)}{2/x^3}\\&=\lim_{x\to\infty}\frac{1}{2+(2/x)}\\&=1/2.\end{align}$$

Answer 2

Hint. Set $t=\dfrac{1}{x}$, and hence $t\to 0$, and $$ x\big((1+\tfrac{1}{x})^x-\mathrm{e}\big)=\frac{(1+t)^{1/t}-\mathrm{e}}{t}=\frac{f(t)-f(0)}{t} =f'(s), $$ for some $s\in (0,t)$ (due to the Mean Value Theorem), where $$ f(t)=\left\{ \begin{array}{cll} (1+t)^{1/t} & \text{if} & t>0,\\ \mathrm{e} & \text{if} & t=0. \end{array} \right. $$ Now $$ f'(t)=\exp\big(\tfrac{1}{t}\ln (1+t)\big)\left(\tfrac{1}{t(1+t)}-\tfrac{\ln(1+ t)}{t^2}\right). $$

So your limit is just $f'(0)$.

In fact, you don't need to find $f'(0)$ directly, but to compute $f'(t)$, for $t>0$ ,and observe that the limit $\lim_{t\to 0} f'(t)$ exists.

Clearly, $\exp\big(\tfrac{1}{t}\ln (1+t)\big)=(1+t)^{1/t}\to\mathrm{e}$, while, for $g(t)=\ln (1+t)$, $$ \frac{1}{t(1+t)}-\frac{\ln(1+ t)}{t^2}=\frac{1}{t}\left(g'(t)-\frac{g(t)-g(0)}{t}\right)\to-\frac{1}{2}g''(0). $$

Answer 3

We have: $$\log\left(1+\frac 1 x\right)^x=x\log\left(1+\frac 1 x\right)=x\left(\frac 1 x-\frac1{2x^2}+o\left(\frac1{x^2}\right)\right)$$ so $$\left(1+\frac 1 x\right)^x=e\exp\left(-\frac1{2x}+o\left(\frac1{x}\right)\right)=e\left(1-\frac1{2x}+o\left(\frac1{x}\right)\right)$$ hence now it's simple to see that the desired limit is: $$-\frac{e}{2}$$

Answer 4

$\displaystyle \begin{aligned}L &= \lim_{x \to \infty}x\left(\left(1 + \frac{1}{x}\right)^{x}- e\right)\\ &= \lim_{x \to \infty}\dfrac{\left(\left(1 + \dfrac{1}{x}\right)^{x}- e\right)}{\dfrac{1}{x}}\\ &= \lim_{y \to 0^{+}}\dfrac{(1 + y)^{1/y} - e}{y}\text{ (by putting }y = 1/x)\\ &= \lim_{y \to 0^{+}}\dfrac{\exp\left(\dfrac{\log(1 + y)}{y}\right) - \exp(1)}{y}\\ &= \lim_{y \to 0^{+}}\dfrac{\exp\left(\dfrac{\log(1 + y)}{y}\right) - \exp(1)}{\dfrac{\log(1 + y)}{y} - 1}\cdot\dfrac{\dfrac{\log(1 + y)}{y} - 1}{y}\\ &= \lim_{z \to 1}\dfrac{\exp(z) - \exp(1)}{z - 1}\cdot\lim_{y \to 0^{+}}\dfrac{\log(1 + y) - y}{y^{2}}\,\,\{\text{by putting }z = (1/y)\log(1 + y)\}\\ &= \exp'(1)\cdot\lim_{y \to 0^{+}}\dfrac{\dfrac{1}{1 + y} - 1}{2y}\text{ (by L'Hospital's Rule)}\\ &= \exp(1)\cdot\lim_{y \to 0^{+}}\frac{-1}{2(1 + y)}\\ &= -\frac{e}{2}\end{aligned}$

One application of LHR is necessary, but the above procedure makes the application of LHR very simple. In the above derivation we have used the fact as $y \to 0^{+}$ the quantity $z = (1/y)\log(1 + y) \to 1$. This limit is pretty standard and a proof can be found in many textbooks and on MSE.