evaluate lim: $\lim _{x\to \infty }\left(2x\left(e-\left(1+\frac{1}{x}\right)^x\right)\right)$

Question

I was trying to solve the above limit and it seems like I'm getting mixed results. I used the fact that: $\lim _{x\to \infty }\left(1+\frac{1}{x}\right)^x = e$

And after that trying with L'Hospital's rule but it didn't get me much further.

Answer 1

HINT: rewrite your term in the form $$\frac{e-\left(1+\frac{1}{x}\right)^{x}}{\frac{1}{2x}}$$ and apply the rules of L'Hospital

Answer 2

With some work, l'Hopital's rule solves the problem.

The hard part is the derivative of $x\mapsto\left(1+\frac1x\right)^x$. For this, write $$y=\left(1+\frac1x\right)^x$$ $$\ln y=x\ln\left(1+\frac1x\right)$$ Now take derivatives: $$\frac{y'}y=\ln\left(1+\frac1x\right)-\frac1{x+1}$$ and hence $$y'=\left(1+\frac1x\right)^x\left[\ln\left(1+\frac1x\right)-\frac1{x+1}\right]$$

Now, l'Hopital:

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

With more l'Hopital and some patience you can find out that

$$\lim_{x\to\infty}x^2\left[\ln\left(1+\frac1x\right)-\frac1{x+1}\right]=\frac12$$

Thus, the final result is $e$.

Warning: I have made all this by hand, so I advise to double check the result.

Answer 3

I'll make use of the following Taylor approximations below: For small $h>0,$ $$\ln (1+h) = h - h^2/2 + O(h^3), e^h = 1 + h +O(h^2).$$

Rewrite our expression as $2xe(1-(1+1/x)^x/e).$ Now

$$\ln[(1+1/x)^x/e] = x\ln (1+1/x) - 1 = x(1/x - 1/2x^2 + O(1/x^3)) - 1 = -1/2x +O(1/x^2).$$

Exponentiating back gives

$$(1+1/x)^x/e = 1-1/2x + O(1/x^2).$$

Thus

$$2xe(1-(1+1/x)^x/e) = 2xe(1/2x + O(1/x^2))=e + O(1/x).$$

So the limit is $e.$

Answer 4

Make the substitution $x \to 1/x$. Thus,

$$\begin{align} \lim_{x\to \infty}\left(2x\left(e-\left(1+\frac1x\right)^x\right)\right)&=\lim_{x\to 0}2\frac{e-\left(1+x\right)^{1/x}}{x}\\\\ &=-2\lim_{x\to 0}\frac{d\left(1+x\right)^{1/x}}{dx}\\\\ &=-2\lim_{x\to 0}\left(1+x\right)^{1/x}\left(\frac{1}{x(x+1)}-\frac{\log(1+x)}{x^2}\right)\\\\ &=e \end{align}$$

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NOTE:

$$\begin{align} \frac{1}{x(x+1)}-\frac{\log(1+x)}{x^2}&=\frac{1}{x(x+1)}-\frac{x-\frac12x^2+O(x^3)}{x^2}\\\\ &=\frac{-1}{x+1}+\frac12 +O(x)\\\\ &\to -\frac12 \end{align}$$

as $x\to 0$.

Answer 5

By standard limits, we have

$$ \frac{{e - \left(1 + \frac{1}{x}\right)^x}}{{\frac{1}{2x}}}= \frac{{e - e^{x\log\left(1 + \frac{1}{x}\right)}}}{{\frac{1}{2x}}}=e\frac{{1 - e^{x\log\left(1 + \frac{1}{x}\right)-1}}}{{\frac{1}{2x}}}=$$

$$=e\cdot \frac{{1 - e^{x\log\left(1 + \frac{1}{x}\right)-1}}}{x\log\left(1 + \frac{1}{x}\right)-1}\cdot 2\frac{\log\left(1 + \frac{1}{x}\right)-\frac1x}{\frac{1}{x^2}} \to e\cdot (-1)\cdot2\left(-\frac12\right)=e$$