$$f(x) = \begin{cases} x^2\sin(1/x), & \text{if $x$ $\neq$ 0} \\ 0, & \text{if $x$ = 0} \end{cases}$$

Ok, as far as I know, to test if a function is continuous at $0$, the $\lim_{x \to 0} \ f(x) = f(0) $ has to exist if $x→0$ , this means that x is different than $0$ , so in this case $f(x)$ = $x^2\sin(1/x)$ , and the limit above exists and equals to $0$. (I think)

How do I derive a piecewise function? I though that i could derive $x^2\sin(1/x) $ after that, $0$. For example, the derivative of

$$f(x) = \begin{cases} x, & \text{if $x$ $≥$ 0} \\ -x, & \text{if $x$ < 0} \end{cases}$$ would be

$$f'(x) = \begin{cases} 1, & \text{if $x$ $≥$ 0} \\ -1, & \text{if $x$ < 0} \end{cases}$$

and $f'(0)$ $= 1$, but this can't be right because the derivative of |x| at $0$ doesn't exist. (i hope I made myself clear with this example)

The exercise asks if the derivative at $0$ exists. Sorry if it caused confusion, i wanna know the derivative at $0$ to the first function and not the $|x|$.

  • 1
    $\begingroup$ You could look at the so called left and right derivatives. If they exist at 0 and are equal then the function is differentiable at 0. $\endgroup$ – BoBoB Apr 21 '17 at 19:53
  • $\begingroup$ For testing differentiability at 0 on the original function you need to fall back to the definition of derivative. $\endgroup$ – Daniel Schepler Apr 21 '17 at 19:59

enter image description here It is the graph of x^2*sin(1/x) and you can draw only one tangent at 0.If a function is differentiable it must has unique target at the desired point. The second function has no derivative at 0 as you see that you can draw two tangents at 0 whose slops are 1 and -1 (you have done it in your question).To make it clear just draw this function and then draw those tangents at 0.


In general when you want to find the derivative of a piece-wise function, you evaluate the two pieces separately, and where they come together, if the function is continuous and the derivative of the left hand side equals the derivative of the right hand side, then you can say that the function is differentiable at that point.

i.e. if $f(x)$ is continous at $a$ and
$\lim_\limits{x\to a^+} \frac {f(x) - f(a)}{x-a} = \lim_\limits{x\to a^-} \frac {f(x) - f(a)}{x-a}$ then f(x) is differentiable at $a.$

To the problem at hand:

$f'(0) = \lim_\limits{x\to 0} \frac {f(x)-f(0)}{x-0}$

$-x^2\le f(x)\le x^2$

$-|x| <\frac {f(x)-f(0)}{x-0} < |x| $

and as $x$ goes to $0, f'(x)$ gets squeezed.


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