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 $e^{x-1}$

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
$$\lim_{x \to 1} \frac{e^{x - 1}}{x^2 - 1} =\left(\lim_{x \to 1} \color{#00bf00}{\frac{1}{x + 1}}\right) \left(\lim_{x \to 1} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}}\right)  = \left(\frac{1}{2}\right)(1) = \frac{1}{2}.$$
