How to evaluate $ \lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1} $ How can you calculate this limit?
$$
\lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1}
$$
I really don't have a clue what to do with the $e^{x-1}$
 A: $e^{x-1}$ is just $e^{x}\cdot e^{-1}=e^x/e$. So factor out $e$ from the limit and you'll be left with 
$$\begin{align*}
\lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1}
&=\dfrac1e\lim_{x \to 1} \frac{e^{x}-e}{x^2-1} \\ 
&=\dfrac1e\color{black}{\lim_{x \to 1} \frac{e^{x}-e^1}{x-1}}\dfrac1{x+1}\\ &=\dfrac1{2e}\color{#C00000}{\lim_{x \to 1} \frac{e^{x}-e^1}{x-1}}.
\end{align*}$$
Now note that by definition of the derivative of $f:x\mapsto e^x$ at $1$: 
$$f'(1)=\color{#C00000}{\lim_{x\to1}\dfrac{e^x-e^1}{x-1}}.$$
A: $u=x-1 \implies \displaystyle\lim_{x\to 1}\dfrac{e^{x-1}-1}{x^2-1}=\displaystyle\lim_{u\to 0}\dfrac{e^{u}-1}{u(u+2)}=\left(\displaystyle\lim_{u\to 0}\dfrac{e^{u}-1}{u}\right)\left(\displaystyle\lim_{u\to 0}\dfrac{1}{u+2}\right)=\dfrac{1}{2}$

Can you figure out the reason behind writing $$\displaystyle\lim_{u\to 0}\dfrac{e^{u}-1}{u(u+2)}=\left(\displaystyle\lim_{u\to 0}\dfrac{e^{u}-1}{u}\right)\left(\displaystyle\lim_{u\to 0}\dfrac{1}{u+2}\right)$$

A: The form of the expression in the limit is something like a difference quotient, and in fact, applying the difference-of-squares factorization $a^2 - b^2 = (a + b)(a - b)$ to the bottom shows that we can write the expression as the product of a difference quotient (whose limit is, by definition, a derivative) and a function continuous at the limit point $x = 1$:
$$\frac{e^{x - 1}}{x^2 - 1} = \color{#00bf00}{\frac{1}{x + 1}} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}}.$$
We can evaluate the limit of this expression by showing that the limits of each of the factors exist and computing them.
The $\color{#bf0000}{\text{red}}$ expression is the difference quotient of the function $$f(x) := e^{x - 1}$$ at $1$, so its limit is
$$\lim_{x \to 1} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}} = f'(1) = e^{x - 1}\vert_{x = 1} = 1.$$
The limit of the $\color{#00bf00}{\text{green}}$ expression is $$\lim_{x \to 1} \color{#00bf00}{\frac{1}{x + 1}} = \frac{1}{2}$$ and so
$$\lim_{x \to 1} \frac{e^{x - 1}}{x^2 - 1} =\left(\lim_{x \to 1} \color{#00bf00}{\frac{1}{x + 1}}\right) \left(\lim_{x \to 1} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}}\right)  = \left(\frac{1}{2}\right)(1) = \frac{1}{2}.$$
