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Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be continuous at $0$.

1.) Prove that $f(x)=xg(x)$ is differentiable at $0$ (via proof form).

2.) Briefly explain how/why the continuity of $g$ at $0$ was needed in part (a).

I'm not sure how.

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What have you tried? What is the definition of differentiable at $0$? (Technically $g$ doesn't need to be continuous at $0$, but at worst it can have a removable discontinuity there.) –  Jonas Meyer Dec 9 '12 at 6:27
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Use the definition of derivative. –  Potato Dec 9 '12 at 6:28
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2 Answers 2

By definition $f$ is differentiable at $0$ if the following limit exists: $$\lim_{x→0}\frac{f(x)-f(0)}{x-0}=\lim_{x→0}\frac{xg(x)-0}{x-0}=\lim_{x→0}g(x)$$

Because $g$ is continuous at $0$ this limits exist and must equal $g(0)$.

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We have $\frac{xg-0}{x-0}=g$. Let the limit approach $0$, you should be able to conclude the limit is $g(0)$. It might be better to try this yourself.

As Jonas Meyer commented, a removable discontinuity is suffice at here.

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Corrected, thanks. –  Bombyx mori Dec 9 '12 at 6:32
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