Why is this function not differentiable at 1 and -1? I'm solving a small calculus problem, which ask me to compute the following limits: 
$$\lim_{x\,\to\,-1^-} \frac{f(x) - f(-1)}{x+1}$$
and 
$$\lim_{x\,\to\,1^+}\frac{f(x) - f(1)}{x-1}.$$
Given that 
$$f(x) = x + \sqrt{x^2 - 1},$$
I get $\infty$ as the result to both limits.
At first look this sounds normal, except after that, in the practicing problem, it says to differentiate the function $f$ on $[-\infty, -1] \cup [1, +\infty]$ which is weird because $f$ is not differentiable at $1$ and $-1$ (from the previously computed limits), so I thought that my limits were wrong, but I couldn't find the mistake.
Please help me out.
Thank's for your time.
 A: Generally we can only compute derivatives on open sets. In this case, if we define the derivative of $f$ on a closed interval $F$ to be it's derivative on the interior of $F$, then there's no problem with the exercise. If, on the other hand, we define the derivative of $f$ on $\pm1$ to be the limit of $f'(x)$ as $x$ approaches $\pm1^{\pm}$, then we can say $f$ is not differentiable on the given set, because none of the above limits exist.
A: You are correct in computing the given limits: for instance
$$
\lim_{x\to 1^+}\frac{f(x)-f(1)}{x-1}=
\lim_{x\to1^+}\frac{x-1+\sqrt{x^2-1}}{x-1}=
\lim_{x\to1^+}\left(1+\sqrt{\frac{x+1}{x-1}}\,\right)=\infty
$$
and the other one is similar.
One can define the derivative at $a$ of a function defined on an interval like $[a,b)$ (where $b$ can be $\infty$) as
$$
\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}
$$
if this limit exists and is finite. In your case the function is not differentiable at $1$ in this sense (unless your textbook allows infinite derivatives, which however is rarely done). It is not really surprising, because the graph of this function is a piece of a hyperbola; if we consider the points of coordinates $(x,y)$ such that $y=x+\sqrt{x^2-1}$, we get
$$
(y-x)^2=x^2-1
$$
that is,
$$
y^2-2xy+1=0
$$

The graph of your function is the part with $y\ge1$ for $x\ge1$ and the part with $-1\le y<0$ for $x\le-1$. As you clearly see, the tangents at $(1,1)$ and $(-1,-1)$ are vertical, which complies with the limits you found.
