Why is $\lim_{x\to 1} (x^2 - 1)/(x-1) = 2$? $$
\lim_{x\to1}\frac{x^2-1}{x-1} = \lim_{x\to1}\frac{(x-1)(x+1)}{x-1} = 2.
$$
I need some more intermediate steps here, please! 
What is happening here? 
The fact that $1-1$ equals $0$ makes it hard for me to understand. 
On the other hand, taking the limit to infinity is a much easier exercise.
 A: The point is: If two functions $f(x)$ and $g(x)$ are equal close to a point, but not necessarily at the point, then they have the same limit.
Let
$$
f(x) = \frac{x^2 - 1}{x - 1}
$$
and
$$
g(x) = x+ 1
$$
Then 
$$
f(x) = g(x)
$$
for all $x\neq 1$. So then
$$
\lim_{x\to 1} f(x) = \lim_{x\to 1} g(x) = \lim_{x\to 1} x + 1 = 2.
$$
A: Hint: $$\frac{(x-1)(x+1)}{(x-1)} = \left(\frac{x-1}{x-1}\right)(x+1)=1\cdot(x+1)$$ Now plug in $x=1$
A: Perhaps your problem is in understanding that $x^2-1=(x-1)(x+1)$.  Multiply out the right side and simplify, and you'll get the left side.
In the fraction $\dfrac{(x-1)(x+1)}{x-1}$, you can cancel a factor from the numerator and the denominator, getting  $\dfrac{x+1}1$.
In the expression you start with, the numerator and the denominator do both approach $0$.
Limits in which the numerator and denominator both approach $0$ but the limit is some finite number like $2$ are immensely more important in calculus than all others sorts of limits combined, because they appear in the expression $\dfrac{dy}{dx}=\lim\limits_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}$, so there could be no derivatives --- thus no differential calculus --- without such limits.
A: for $$x\ne 1$$ we have $$\frac{(x-1)(x+1)}{x-1}=x+1$$
A: I can only think of one step to add:
$$
\lim_{x\to 1}\frac{x^2-1}{x-1}
=\lim_{x\to 1}\frac{(x-1)(x+1)}{x-1}
=\lim_{x\to 1}x+1
=2
$$
A: From the definition:
$$\left| \frac{x^2-1}{x-1} - 2 \right| < \epsilon$$
For some $\delta$ such that: $|x - 1| < \delta$
When we simply the first equation I wrote here you get:
$$\left| x - 1  \right| < \epsilon$$
Because we have to find a $\delta$  so that: $|x - 1| < \delta$
We can let $\delta = \epsilon$
Therefore, we found a $\delta$ so that the proof is sufficed. Which is why the limit is true. $\blacksquare$
