Help me with this limit

Question

$$ \lim_{x\to0} {{xe^x \over e^x-1}-1 \over x}$$

I know it should equal ${1 \over 2}$ because when i calculate with number like $0.0001$ the limit $\approx {1 \over 2}$ but i can't prove it.

Answer 1

First multiply numerator and denominator by $e^x-1$ to rewrite the fraction as

$$\frac{xe^x - e^x + 1}{xe^x-x}$$

Solution 1:

Now we apply L'Hôpital's rule twice to see

$$\lim_{x\rightarrow 0}\frac{xe^x - e^x + 1}{xe^x-x}=\lim_{x\rightarrow 0}\frac{xe^x}{e^x+xe^x-1}=\lim_{x\rightarrow 0} \frac{e^x+xe^x}{2e^x+xe^x}=\frac{1}{2}$$

Solution 2: (without L'Hôpital's rule)

Remember that $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. That is, we have $e^x=1+x+\frac{1}{2}x^2+O(x^3)$ close to $0$. Therefore

$$\frac{xe^x - e^x + 1}{xe^x-x}=\frac{(1+x+\frac{1}{2}x^2+O(x^3))(x-1)+1}{x(x+\frac{1}{2}x^2+O(x^3))}=\frac{\frac{1}{2}x^2+O(x^3)}{x^2+O(x^3)}=\frac{1+O(x)}{2+O(x)}\rightarrow\frac{1}{2}$$

as $x\rightarrow 0$.

In case you're not familiar with the $O$-notation:

$O(x^n)$ is just a placeholder for some function $f$ such there exists some $C>0$ and $x_0>0$ such that $|f(x)|\le C|x^n|$ for all $x$ with $|x|\le x_0$, i.e. $O(x^n)$ means asymptotic growth at most like $x^n$ near $0$.

Answer 2

Divide the top and bottom by $x$ to clean stuff up:

$$\dots={ {{e^x \over e^x-1}-\frac{1}{x} }}\normalsize=\frac{x\cdot e^x-e^x+1}{x\cdot(e^x-1)}$$

Can you do it now?

Answer 3

Ok, I there, what you do is use L'Hosptial several times:

Your limit is a limit of the type $\frac{0}{0}$. Using L'Hospital yields: $$\lim_{x\rightarrow 0}(\ldots) = \lim_{x\rightarrow 0}\frac{\exp(2x)-\exp(x)-x\exp(x)}{(\exp(x)-1)^2}$$ Then using L'Hospital twice more (in each case you will still have limit of the type $\frac{0}{0}$) yields: $$\lim_{x\rightarrow 0}(\ldots) = \lim_{x\rightarrow 0} \frac{4\exp{2x}-3\exp(x)-x\exp(x)}{2\exp{x}}$$ And using $\lim_{x\rightarrow 0} \exp(x) = 1$ then yiels the result of $\frac{1}{2}$ you were looking for! Hope that helps :).

I'm reading above you can for some reason not use L'Hospital. Alternatively you can use the Taylor expansion for the exponential function: $$\exp(x) = 1 + x + \frac{x^2}{2} + \ldots$$ Then after rearranging the expression as the smart people above me did: $$\frac{x\exp(x)-\exp(x)+1}{x\exp(x)-x} = \frac{x+x^2+\mathcal{O}(x^2) - 1 - x - 0.5\cdot x^2 \mathcal{O}(x^2) + 1}{x+x^2+\mathcal{O}(x^2)-x}$$ Rearranging gives: $$\lim_{x\rightarrow 0} (\ldots) = \lim_{x\rightarrow 0} \frac{\frac{x^2}{2}+\mathcal{O}(x^2)}{x^2+\mathcal{O}(x^2)} = \frac{1}{2}$$