how a sequence relates to a derivative?

**bkbowser** · November 14th 2012, 09:30 PM

i'm having some trouble with the ideas behind this question.
f is a continuous, defferentiable function with f(0)=0
given $a_{n}=nf(\frac{1}{n})$ prove that $\lim_{x\to\infty}=f^\prime(0)$

so i need a way to relate the value that some sequence converges to and the first derivative of a related function evaluated at 0.

i'm having a really hard time with this question. effectively i've got nothing on it and its been a good 5-7 hours so i don't think i can solve it with what i know already.

**MarkFL** · November 14th 2012, 09:45 PM

$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{f\left (\frac{1}{n} \right)}{\frac{1}{n}}$

This is the indeterminate form 0/0, so apply L'Hôpital's rule to get the desired result.

**bkbowser** · November 14th 2012, 10:00 PM

so the numerator would be $f^\prime(1/n)*(1/n)^\prime$ by the chain rule

and the denominator $f(1/n)^\prime$

so then i have $\lim_{x\to\infty} a_n = \lim_{x\to\infty} f^\prime(1/n)$

and since 1/n goes to 0 then it becomes

$\lim_{x\to\infty} a_n = \lim_{x\to\infty} f^\prime(0)$

and i'm done?

**MarkFL** · November 14th 2012, 10:05 PM

Your numerator is correct, but the denominator would simply be (1/n)', which cancels with this same factor in the numerator that results from the chain rule.

After that, what you wrote is correct, except let n go to infinity rather than x.

**bkbowser** · November 14th 2012, 10:07 PM

kk thanks.

**bkbowser** · November 14th 2012, 10:12 PM

$f^\prime(1/n)*(1/n)^\prime$ by the chain rule

and the denominator $(1/n)^\prime$

so then i have $\lim_{n\to\infty} a_n = \lim_{n\to\infty} f^\prime(1/n)$

and since 1/n goes to 0 then it becomes

$\lim_{n\to\infty} a_n = f^\prime(0)$

and i'm done?

so then this? sorry im exhausted.

**MarkFL** · November 14th 2012, 10:17 PM

Yes, that looks good!

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