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I am redoing my math education after 35 years in order to be up to date with my children's curriculum (I usually have a nanosecond to give an answer because of magical dad knowledge). My older son is in high school and they are going through derivatives.

One of the exercises I was preparing for him (this is beyond the homework, he is in an "advanced" track and I want to show him some more interesting problems) I stumbled upon the following problem. It is a translation from French so feel free to edit in order to improve the wording

We define

$$ \begin{align} f: \mathbb{R} & \rightarrow \mathbb{R} \\ x & \mapsto \begin{cases} \text{$x^2 - 1$ if $x < 0$} \\ \text{$x^2 + 1$ if $x \geq 0$} \end{cases} \end{align} $$

Demonstrate that, for $a=0$, $f$ is derivable on the right but not on the left.

Both $x^2 - 1$ and $x^2 + 1$ are derivable in $\mathbb{R}$ but $f$ is defined for $0$ as $x^2 + 1$. So technically the derivative in $0$ will be $2 \times 0 = 0$

It is however derivable on $0^+$ ($2x$) and on $0^-$ ($2x$).

The only idea I have is that by taking the definition of the left derivative I have $$ \lim_{h \rightarrow 0^-} \frac{f(x+h) - f(x)}{h} $$ which I think is not defined because the limit approaching on the left does not exist (as the function is undefined there). But I think I must be missing something because this is too obvious.

WoJ
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  • This isn't a hard question, but you are somewhat oversimplifying. Why does the limit not existing imply that THIS limit doesn't exist? – Rushabh Mehta Dec 30 '20 at 18:41
  • @DonThousand: sorry, I do not understand your question (which is certainly the right one, it is just that I do not understand it) – WoJ Dec 30 '20 at 18:44
  • you say that this limit (the limit of the left derivative) doesn't exist because the limit of the function doesn't exist (which, FYI, it does). That doesn't make sense/ – Rushabh Mehta Dec 30 '20 at 18:44

2 Answers2

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You are on the right track, but as I note in a comment, you are somewhat oversimplifying.

We do want to use the definitions of left/right derivative, so let's do that.

$$\lim\limits_{h\to0^-}\frac{f(x+h)-f(x)}h=\lim\limits_{h\to0^-}\frac{(0+h)^2-1-(0^2+1)}h=\lim\limits_{h\to0^-}\frac{h^2-2}h=\infty$$

$$\lim\limits_{h\to0^+}\frac{f(x+h)-f(x)}h=\lim\limits_{h\to0^+}\frac{(0+h)^2+1-(0^2+1)}h=\lim\limits_{h\to0^+}h=0$$

There is no such thing as an infinite derivative, so this shows that only the right derivative is well defined at $0$.

Rushabh Mehta
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  • ah crap, yes of course, it would have enough for me to do the next step and actually calculate the limit. Thanks a lot! – WoJ Dec 30 '20 at 19:17
  • Now that I think about it, your first limit with $x \in \mathbb{R}$ (instead of the discontinuous function) is the derivative for $x=0$. Which is still $\inf$, despite existing there (this is a smooth parabola with its extremum at $x=0$). I will think about that a bit more. – WoJ Dec 30 '20 at 19:32
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To determine whether or not $f'(c)$ exists, one can see if the following is true:

$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}.$$

If either the LHS or RHS to the above equation does not exist, or the LHS and RHS exists but have different values, then $f'(c)$ does not exist.

For your piecewise function, let's determine whether or not the derivative of $f$ at $c = 0$ exists.

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

However,

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

or rather it's more proper to say that the latter limit does not exist.

Since the second limit does not exist, $f'(0)$ does not exist.

Adam Rubinson
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