Proving a continuous function isn't necessarily differentiable.

I have been advised by my lecturer to use $|x|$ as my function example to prove this. I have also read on a thread here that this function is continuous for all values but not differentiable at $0$ (so the limit does not exist here). According to my notes, this can be proven by computing the limits either side of $0$, and this is where I am struggling. How should I go about doing this? Any help would be greatly appreciated.

  • $\begingroup$ Write down the definition of the derivative from the right and from the left. What part are you having trouble with, beyond that? $\endgroup$ – symplectomorphic Nov 18 '17 at 18:16
  • $\begingroup$ Recall the definition for $|x|$ and how it is defined for positive $x$ and negative $x$, separately. Then use that definition to calculate the left-hand derivative and right-hand derivative separately. These limits are not equal. $\endgroup$ – Clayton Nov 18 '17 at 18:19
  • $\begingroup$ Have you drawn a diagram - it should then be obvious what is happening. $\endgroup$ – Mark Bennet Nov 18 '17 at 18:25
  • $\begingroup$ Compute $\lim_{x \uparrow 0} {|x|-|0| \over x}$ and $\lim_{x \downarrow 0} {|x|-|0| \over x}$. If the function was differentiable at $x=0$ then these two values would be the same. $\endgroup$ – copper.hat Nov 18 '17 at 18:27
  • $\begingroup$ I am confused with why we must put the mod 0 in the working. Could you please explain this? Aside from that, does what you have said mean the limits are 1 and -1? $\endgroup$ – T.meeley Nov 18 '17 at 18:33

$$\lim_{x \to 0^-} \frac{|x|-|0|}{ x}$$ Limit as $x\to 0^-$ means that $x<0$ and then $|x|=-x$ so the limit becomes $$\lim_{x \to 0^-} \frac{-x}{ x}=-1$$ In a similar way $$\lim_{x \to 0^+} \frac{+x}{ x}=+1$$ As the two limits are different the limit doesn't exist and $f(x)=|x|$ is not differentiable at $x=0$.

Hope this helps

  • $\begingroup$ It does thankyou! I ended up writing something like that so I’m glad I was along the right lines. $\endgroup$ – T.meeley Nov 19 '17 at 22:31

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