I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\frac{e^x-e^{-x}}{e^x-1}$$
For $x\rightarrow0$, the denominator is asymptotic to
$$e^x-1\sim x$$
Here's how I simplify the numerator:
$$e^x-e^{-x}\sim 1-e^{-x}=-(e^{-x}-1)\sim-(-x)=x$$
Finally we have
$$f(x)\sim\frac{x}{x}=1$$
My textbook does it another way and the solution is 2. I wonder what I'm doing wrong. Perhaps it's the fact that I have replaced $e^x$ with $1$ but that seemed to make sense to me because $e^x\rightarrow1$ for $x\rightarrow0$. Any hints?
Best Answer
What you did is wrong because both $e^x$ and $e^{-x}$ behave as $1$ near $0$. You can't replace one if them by $1$, while the other one remains as it is.
Note that$$\lim_{x\to0}\frac{e^x-e^{-x}}{e^x-1}=\lim_{x\to0}\frac{e^x+e^{-x}}{e^x}=2.$$