How to prove that a function is continuous? Could you give me some hint how to solve this question:
Suppose $f$ is a differentiable function for all $0<x<1$,$f(0)=1,f'(x)>0$ in the given interval.
It is obvious that $f$ is continuous for all $0<x<1$, but is it continuous at $x=0$ ?
Thanks.
 A: No and the example is very simple:
$$
f(x)=\begin{cases}
1 & \text{if $x=0$}\\
x & \text{if $0<x<1$}
\end{cases}
$$
The function is differentiable in the open interval $(0,1)$ and its derivative is $f'(x)=1>0$ for all $x\in(0,1)$. However,
$$
\lim_{x\to0}f(x)=0\ne1=f(0)
$$
so $f$ is not continuous at $0$.
Differentiability at every point of $(0,1)$ can't imply anything about the behavior of the function at $0$, because for each $x>0$ there is $\delta>0$ such that $(x-\delta,x+\delta)\subset(0,1)$ and the differentiability of $f$ at $x$ only involves the values of $f$ in $(x-\delta,x+\delta)$, which $0$ doesn't belong to.
A: First, knowing the definition of a continuous function helps.
Quoting wikipedia here,
A function f(x) is continuous at point c if the limit of f(x) as x approaches c is f(c).
You have some facts stated earlier:


*

*f(x) is differentiable in 0 < x < 1

*f(0) = 1

*f'(x)>0 in (0,1)


We should check the limit of f(x) as x approaches 0. The answer is going to be a number, whose derivative is >0. However, Since f'(0) = 0, we can automatically assume that 
lim as f(x) approaches 0 isn't equal to f(0) because their derivatives are different. 
Hence f(x) is not continuous at 0. 
Hope this makes sense.
