Proving constants are differentiable using the definition I'm currently studying differentiability at college, and I can't for the life of me understand how a constant is differentiable. I am using the definition of differentiability (as $h \rightarrow 0$ and you have $h$ at the bottom).
When $f(x) =c$ what is $f(x_0 + h)?$ 
I understand why x is differentiable but I can't get a hold for constants! 
Thank you for reading!
 A: Notice that a constant function $f(x) = c$ $\forall x$. That is: 
\begin{equation}
f(x_0 + h) = c = f(x_0)
\end{equation}
Using the definition of differentiability: 
\begin{equation}
\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = \lim_{h \to 0}\frac{c - c}{h} =\lim_{h \to 0} 0 = 0
\end{equation}
A: As far as I understand, your problem lies a couple of lessons further back, so I'll start there:
Limits
What is a limit of a function? By definition, we have that $$\lim_{x\to a} f(x) = L$$
if and only if for every $\epsilon>0,$ there exists such a $\delta > 0$ that for all $x$ such that $0<|x-a|<\delta$, we have $|L-f(x)|<\epsilon$
What does this mean? Well, it means that the values of $f(x)$ are close to $L$ if $x$ is close to $a$. But there is an important inequality there: $0<|x-a|$. Yes, $x$ is close to $a$, but never equal to it. So, for example, the function $f(x)=\frac {x^2}{x}$, which is undefined at $x=0$, may still have a limit at $x=0$. In fact, that limit is $0$, because:


*

*Take any $\epsilon>0$

*Set $\delta = \epsilon$, and let $x$ be such that $|x-a| < \delta$ (where $a$ is of course $0$)

*Then, we have $|f(x)-0|=|\frac{x^2}{x}-0| = |x-0|=|x|<\epsilon$, and we are done


Note: the equality $|\frac{x^2}{x}-0| = |x-0|$ **only holds if $x\neq 0$, but the limit does not care about $x=0$, so we are safe.
