Limit of a 0/0 function Let's say we have a function, for example,
$$
f(x) = \frac{x-1}{x^2+2x-3},
$$
and we want to now what is 
$$
\lim_{x \to 1} f(x).
$$
The result is $\frac{1}{4}$.
So there exists a limit as $x \to 1$.
My teacher says that the limit at $x=1$ doesn't exist. How is that? I don't understand it. We know that a limit exists when the one sided limits are the same result.
Thank you!

It's possible that your teacher was pointing out the fact that the function doesn't exist at $x = 1$.  That's different from saying that the limit doesn't exist as $x \to 1$.  Notice that by factoring,
$$
f(x) = \frac{x-1}{x^2 + 2x - 3} = \frac{x-1}{(x-1)(x+3)}
$$
As long as we are considering $x \ne 1$, the last expression simplifies:
$$
\frac{x-1}{(x-1)(x+3)} = \frac{1}{x+3}.
$$
In other words, for any $x$ other than exactly $1$,
$$
f(x) = \frac{1}{x+3}.
$$
This helps understand what happens as $x$ gets ever closer to $1$:  $f(x)$ gets ever closer to
$$
\frac{1}{(1) + 3} = \frac{1}{4}.
$$
