# Thread: lim x->0 [(e^x-1)/x]

1. ## lim x->0 [(e^x-1)/x]

Hi, I have been trying to solve this limit for past 4 hours, and I can't figure it out. Can anyone guide me ?

lim x->0 [1/x - 1/((e^x)-1)]

without L'Hosptial, without boundaries
it should be: 1/2; use lim x->0 [((e^x) -1)/x = 1]

sorry for wrong topic name, I messed it up, I'm new here.

2. Originally Posted by MarsTeam
Hi, I have been trying to solve this limit for past 4 hours, and I can't figure it out. Can anyone guide me ?

lim x->0 [1/x - 1/((e^x)-1)]

without L'Hosptial, without boundaries
it should be: 1/2; use lim x->0 [((e^x) -1)/x = 1]

sorry for wrong topic name, I messed it up, I'm new here.
Problem: Compute $\displaystyle L=\lim_{x\to0}\left\{\frac{1}{x}-\frac{1}{e^x-1}\right\}$

Solution: Note that $\displaystyle \frac{1}{x}-\frac{1}{e^x-1}=\frac{e^x-1-x}{x\left(e^x-1\right)}$. Now I am not exactly sure what boundaries are (it may be this) but $\displaystyle e^x-1\stackrel{0}{\sim}x$ and $\displaystyle e^x-x-1\stackrel{0}{\sim}1+x+\frac{x^2}{2}-1-x=\frac{x^2}{2}$ so that $\displaystyle L=\lim_{x\to0}\left\{\frac{1}{x}-\frac{1}{e^x-1}\right\}=\lim_{x\to0}\frac{e^x-x-1}{x(e^x-1)}=\lim_{x\to0}\frac{\tfrac{x^2}{2}}{x\cdot x}=\frac{1}{2}$.

Note that if you are weary about using asymptotic equivalences. Merely note that $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. A particular implications of this is that if $\displaystyle |x|<1$ we see $\displaystyle x+\frac{x^2}\le e^x\-1\le x+\frac{x^2}{2}+\frac{2x^3}{3!}$ for sufficiently small $\displaystyle x$.