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Let's say I have a particular function, $f(x)$.

I want to check whether f is differentiable at $x = a$, I try to do this by using the formal definition of the derivative in terms of a limit, i.e.

$$f^\prime (a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$$

and suppose when I evaluate the limit w.r.t. $h$, I have when $h$ tends to $0$ from below that the limit is $-b$, and when $h$ tends to $0$ from above the limit is $b$.

I am confused at this, obviously I think that this means that $f$ is not differentiable at $x = a$, however, in examples I have been given of a question of this sort, the solutions evaluate the modulus function of $f^\prime (x)$ and not just $f^\prime (x)$ (there are some questions where it doesn't). Why is this used? And most importantly, it is relevant to my situation, as in all of these examples they end up with the limit tending to $0$, which is obviously the only non-positive real number not affected by the modulus function. Would I just show that $x = a$ is not differentiable because the limit of $h$ tends to $0$ from below is equal to $-b$ and the limit of $h$ tends to $0$ from above is equal to $b$ and they are not equal? What is the relevance of the modulus function in this?

  • Please use MathJax. – saulspatz May 26 '20 at 23:01
  • Your $f$ is not differentiable at $0$. We cannot tell why the solutions you've been given evaluate $|f^\prime|$ without knowing the problems. – Pacciu May 26 '20 at 23:03
  • One of the examples is f(x) = |x|^5/4 , in the solutions I've been given they use | |h|^5/4 / h | = |h|^1/4 which tends to 0 as h tends to 0, thus f is differentiable at x = 0. . –  May 26 '20 at 23:08

1 Answers1

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If $$\lim_{h\to 0}\Bigl|\frac{f(x+h)-f(x)}{h}\Bigr|=L(>0)$$

then we can say nothing about $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

but if $L=0$, we conclude that $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=0$$ and that $ f$ is differentiable at $x$ and $$f'(x)=0$$

hamam_Abdallah
  • 62,951
  • So if, say, L = 1, then a sufficient proof that f is not differentiable at x would be to prove that the limit as h tends to 0 from below is not equal to the limits as h tends to 0 from above –  May 26 '20 at 23:16
  • Yes . the two limits must be equal, the function to be differentiable. – hamam_Abdallah May 26 '20 at 23:18