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$f(x) = \begin{cases}x^4\left(2 + \sin\frac1x\right) & x\neq 0\\ 0 & x=0.\end{cases}$

Prove that $f$ is differentiable on $\Bbb R$.

I know how to prove that $f$ is differentiable at $x = 0$ by showing that $\lim_{h \to 0} \frac {f(0 + h) - f(0)}h = 0$ and solving it, but how do i prove that it is differentiable if $x$ doesnt equal 0? Is it showing $\lim_{h \to 0} \frac {f(x + h) - f(x)}h?$ Stuck can someone show me how?

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    "I know how to prove that $f$ is differentiable at $x=0$ by showing that $\lim_{h\to0}\frac{f(0+h)−f(0)}{h}=0$ and solving it [...]" — FYI this is a common misuse of language. One doesn't solve a limit, one computes it. Essentially your "and solving it" is not only unnecessary, it's also gibberish.2012-11-06

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$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}x^3\left(2+\sin\frac{1}{x}\right)=0$

since $\,2+\sin\dfrac{1}{x}\,$ is bounded.

Added: If $\,x\neq 0\,$ the $\,f(x)\,$ is the product of two differentiable functions and thus differentiable itself. You can apply the product rule + the chain rule:

$x\neq 0\Longrightarrow f'(x)=4x^3\left(2+\sin\frac{1}{x}\right)+x^4\left(-\frac{1}{x^2}\cos\frac{1}{x}\right)=$

$=x^2\left(8x+4x\sin\frac{1}{x}-\cos\frac{1}{x}\right)$

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    Thanx @wj32, I added a complete answer to my post above.2012-11-06
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$ \lim_{h \to 0} \frac {f(x + h) - f(x)}h =\lim_{h \to 0} \frac{ (x+h)^4\left(2 + \sin\frac {1}{x+h}\right)-x^4\left(2 + \sin\frac1x\right) }{h}\,.$

Now, to evaluate the above limit, note that, $(x+h)^4 = x^4+4x^3h+6x^2h^2+4xh^3+h^4 $. I think, it is easy now to evaluate the limit.

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    Is there some advantage to applying l'Hôpital's rule rather than using the product rule and chain rule? In order to apply l'Hôpital's rule, you still have to take the derivative of the numerator, which seems to me to be essentially the same problem all over again. If $g(h) = f(x+h)$, then $g'(0)=f'(x)$.2012-11-06