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$f$ is differentiable. Prove that if $\lim_{n \rightarrow \infty}n(f(\frac{1}{n})-f(0))=L$ then $f'(0)=L$.

I tried L'Hopital: $\lim_{n \rightarrow \infty}n(f(\frac{1}{n})-f(0))=\lim_{n \rightarrow \infty} \frac {f(\frac{1}{n})-f(0)}{\frac{1}{n}}$, but it didn't get me far...

Any assistance would be much appreciated!

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  • $\begingroup$ $f$ is differentiable. $\endgroup$ – Reyo Feb 8 '15 at 18:34
  • $\begingroup$ Let $h = \frac{1}{n}$ in your limit $\endgroup$ – Zach Effman Feb 8 '15 at 18:36
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Didn't you solve it already?

"I tried L'Hopital: $lim_{n→∞}n(f(\frac{1}{n})−f(0))=lim_{n→∞}\frac{f(\frac{1}{n})−f(0)}{\frac{1}{n}}$, but it didn't get me far."

Isn't the definition of a derivative by using limits:

$lim_{h→0}\frac{f(x+h)−f(x)}{h}$

So for your case, just let $h = \frac{1}{n}$

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  • $\begingroup$ yeah, didn't notice something so trivial! $\endgroup$ – Reyo Feb 8 '15 at 18:44

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