Tricky limit which probably requires L'Hopital

Question

$$\lim_{h\to0}\left\{\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{(1-e^h)^3}\right\}$$

I arrived at this limit when evaluating some residues. Wolfram Alpha tells me it is equal to $-1/6$. But this is such a complicated expression, how would one go about working out the limit? Since each term in the numerator goes to $0$ at the limit, I separated them and tried to use L'Hopital, but this does not work because we get something like $\frac{finite}0-\frac{finite}0$ which is basically $\infty-\infty$. This is not helpful to me.

How can I do it?

Answer 1

By doing what you did, you have essentially seen that you can't take the limits separately - take them together and you'll get another expression on the numerator which still goes to $0$. Then apply L'Hopital twice more and you're there.

After the first L'hopital application you should get $$\lim_{h\to0}\left\{\frac{(1+h)(e^h+1)+he^h-4e^h+2}{-3(1-e^t)^2}\right\}$$Notice the numerator goes to $2+0-4+2=0$ at the limit, as you want.

Answer 2

The Taylor series of the numerator at $0$ begins with $\frac{h^3}6$, whereas the Taylor series of the denominator at $0$ begins with $-h^3$ (obviously). Therefore, the limit at $0$ is $-\frac{1}6$.

Answer 3

Note that $$\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{(1-e^h)^3}=\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{ h^3 }\frac{h^3}{(1-e^h)^3} =e^h\frac{he^{h}+h+2-2e^h}{ h^3 }\left(\frac{h}{1-e^h}\right)^3\to1\cdot \frac 16\cdot(-1)=-\frac16$$

indeed by l'Hopital

$$\lim_{h\to 0}\frac{he^{h}+h+2-2e^h}{ h^3 }=\lim_{h\to 0}\frac{he^h+e^h+1-2e^h}{3h^2}=\lim_{h\to 0}\frac{he^h+e^h-e^h}{6h}=\lim_{h\to 0}\frac{e^h}{6}=\frac16$$