What is $\lim_{x \to 0} \left(\frac{1}{e^x-1}-\frac{1}{x}\right)$ without l'Hopital's rule? [duplicate]

Question

Any ideas on how to calculate $\lim_{x \to 0} \left(\frac{1}{e^x-1}-\frac{1}{x}\right)$ without using l'Hopital's rule?

I tried putting $u = e^x-1$ and $x = \ln(u+1)$, replacing but i dont get much farther than that. I wanted to use the limit $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$ along to solve it but dont know how.

$$\lim_{x \to 0} \left(\frac{1}{e^x-1}-\frac{1}{x}\right) = \lim_{u \to 0} \left(\frac{1}{u}-\frac{1}{\ln(u+1)}\right) = \lim_{u \to 0} \left(\frac{\ln(u+1)-u}{u\cdot \ln(u+1)}\right) $$

If the $u$ multiplying in the denominator were to be $ \frac{1}{u} $, I could make $\frac{1}{u} \cdot \ln(u+1) = \ln(u+1)^\frac{1}{u}$ and do $t = \frac{1}{u} $ (with $ {u \to 0} $ implying $ {t \to \infty} $) and $ \ln(u+1)^\frac{1}{u} = \ln(\frac{1}{t} + 1)^t $. Then $ \lim_{t \to \infty} \ln(\frac{1}{t} + 1)^t = \ln(\lim_{t \to \infty} \left(\frac{1}{t} + 1)^t\right) = \ln(e) = 1 $. That is as far as I got.

Answer 1

The substitution you tried can't help so much, instead we can try with

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

which reduces to the standard limit $\lim_{x\to 0}\frac{e^x-1}x$ and to

$$\lim_{x\to 0}\frac{e^x-x-1}{x^2}$$

which is discussed here

• Are all limits solvable without L'Hôpital Rule or Series Expansion

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Note that the way you were looking for by substitution leads to a similar solution indeed from here

$$\ldots=\frac{\ln(u+1)-u}{u\cdot \ln(u+1)}=\frac{u}{\ln (1+u)}\frac{\ln(1+u)-u}{u^2}$$

which reduces to the standard limit $\lim_{x\to 0}\frac{\ln(1+x)}x$ and to

$$\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}$$

which is discussed in the same reference.

Answer 2

Asymptotics ... As $x \to 0$, \begin{align} e^x &= 1+x+\frac{1}{2}x^2 + o(x^2) \\ e^x-1 &= x+\frac{1}{2}x^2+o(x^2) = x\left(1+\frac{1}{2}x + o(x)\right) \\ \frac{1}{e^x-1} &= \frac{1}{x}(1-\frac{1}{2}x + o(x)) = \frac{1}{x}-\frac{1}{2} + o(1) \\ \frac{1}{e^x-1} - \frac{1}{x} &= -\frac{1}{2} + o(1) \\ \lim_{x\to 0}\left(\frac{1}{e^x-1} - \frac{1}{x}\right) &= -\frac{1}{2} \end{align}