Is it true that the function f(x) goes to 0 as n goes to infinity

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.

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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:

1. $\lim_{x\rightarrow 0+}f(x)$ exists.

2. 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.

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Hi Vadim, well, it is provided that f is continuous on [0,1] and differentiable on (0,1) and we just want to prove if it is true that f goes 0 when x goes to 0 from above. – LoveMath Apr 27 '13 at 0:40

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.

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