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 $0&lt;x&lt;1$,$f(0)=1,f'(x)&gt;0$ in the given interval.
It is obvious that $f$ is continuous for all $0&lt;x&lt;1$, but is it continuous at $x=0$ ?

Thanks.
No and the example is very simple:
$$
f(x)=\begin{cases}
1 &amp; \text{if $x=0$}\\
x &amp; \text{if $0&lt;x&lt;1$}
\end{cases}
$$
The function is differentiable in the open interval $(0,1)$ and its derivative is $f'(x)=1&gt;0$ 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 $x&gt;0$ there is $\delta&gt;0$ 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.
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 &lt; x &lt; 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.