# Show that a function is differentiable

I have a function $$f(x)= \begin{cases} x & \text{if } x \ge0 \\ x^2 & \text{if } x<0 \end{cases}$$ and want to show that it is continuous but not differentiable at $x=0$

Now to show that a function is differentable we show that $$f'(x_0)= \lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

but I am always confused with such funtions. Do I have to choose $x$ or $x^2$

Taking the comments into consideration a functions is differentiable if the difference quotient $\frac{f(x)-f(x_0)}{x-x_0}, x_0\ne0$ approaches a limit. And limit exist only if left- and right-hand side limit is equal. So $$\lim_{x \to 0^-}= \frac{x^2-0}{x-0}=\frac{x^2}{x}=x=0$$ and $$\lim_{x \to 0^+}= \frac{x-0}{x-0}=\frac{x}{x}=1$$ thus they are not equal which means f is not differentiable.

For the continuity part I am considering the definition: $\forall \varepsilon >0 \ \exists \delta>0$ s.t $\mid f(x)-f(x_0)\mid < \varepsilon$ if $\mid x-x_0 \mid < \delta.$

• Find $f'$ in two cases. Jan 29, 2017 at 15:58
• You want to show it's NOT differentiable, i.e., that this limit does not exist. Do you know the theorem that says that the limit exists only if the left and right limits exist and are equal? Assuming so, when computing the right limit ($x$ goes to zero from above), you know that you're computing $f(x)$ for $x > 0$, so you use the $x$ part of the definition. For the left lim, you use the $x^2$ part of the definition. Give that a try, edit your question to show where you get, and we can help you further perhaps if you can't get there yourself. Jan 29, 2017 at 15:58
• I have edited my question :) Jan 29, 2017 at 16:21

You need to show that this limit does not exist. In order to so, since your function has a different form right and left of 0, take the limits as $x \to 0^+$ and $x \to 0^-$ respectively:
$$\lim_{x \to 0^+}\frac{f(x) - f(0)}{x - 0}$$
$$\lim_{x \to 0^-}\frac{f(x) - f(0)}{x - 0}$$
Computing the one sided limits is easier, because you know that when $x$ approaches 0 from positive $x$, $f$ is equal to $x$, but when it approaches 0 from negative $x$, $f$ is equal to $x^2$.
Just take the left and right limits $$\lim_{x\to 0^+}\frac{x-0}{x-0}=1\ne 0=\lim_{x\to x 0^-}\frac{x^2-0}{x-0}$$