Suppose that a function $f(x)$ defined on $[0,1]$ satisfies $f(1/n)\to 0$ as $n\to\infty$. Is it true that $f(x)\to 0$ as $x\to 0^+$ provided
(a) $f$ is continuous on $[0,1]$ ?
(b) $f$ is differentiable $(0,1)$ ?
I know it will be true for example for the function $f(x)=sin(x)$ but is it true for any function $f(x)$? and how to prove it? I am thinking if $\frac{1}{n}$=$x$ then $x=\frac{1}{n}$ but I don't know how continue?
Can any one solve this problem for me?
Thanks so much.
First, if a function is differentiable at a point then it is continuous at that point, so if (a) is true then (b) is also true.
Second, to answer (a), you need to prove two things, assuming $f(x)$ is continuous:
$\lim_{x\rightarrow 0+}f(x)$ exists.
Assuming it exists, $\lim_{x\rightarrow 0+}f(x)=0$.
Perhaps you should try proving each of these to see if you still need more help.
If these are two separate questions, it is true that differentiable implies continuous, but you don't have differentiable st zero. Let $f(x)=0$ for $x \neq 0, 1$ for $x=0$.
For the first, if somebody claims that $f(0)=L\neq 0$ you can use $\epsilon=\frac L2$ to prove the function is not continuous.
