# Non-Differentiable Function proof

• member 731016
member 731016
Homework Statement
Relevant Equations
For this problem,

I am trying to prove that this function is non-differentiable at 0.

In order for a function to be non-differentiable at zero, then the derivative must not exist at zero ##⇔ \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}## does not exist or ##⇔ \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \neq \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}##

##\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{x + 1 - 1}{x} = 1##

##\lim_{x \to 0^+} \frac{\cos x - \cos 0}{x} = \lim_{x \to 0^+} \frac{\cos x - 1}{x}## Howver, I'm not sure where to go from here.

Thanks!

Hint: ##\cos x \leq 1##.

member 731016 and Office_Shredder
Both numerator and denominator in your last limit approach zero. Use the l'Hopital rule!

Hill said:
Hint: ##\cos x \leq 1##.
This does not really help finding the correct limit. ##1-x## is also ##\leq 1## for ##x > 0## and gives a different limit than ##\cos(x)##.

member 731016
Orodruin said:
This does not really help finding the correct limit.
Of course it does not, but the OP does not need it. It only needs to show that this limit is not ##1##.

Last edited:
member 731016
Hey OP. The proof rests on the fact that if the left sided and right sided limits at ##x=0## aren't equal, then the limit at ##x=0## is undefined. You're almost there, just a bit more work is needed.

member 731016
Hill said:
Of course it does not, but the OP does not need it. It only needs to show that this limit is not ##1##.
This is true, but not that much easier than to actually compute the limit using the l'Hopital rule.

member 731016
Orodruin said:
This is true, but not that much easier than to actually compute the limit using the l'Hopital rule.

I find it a bit weird to suggest someone compute the derivative of cos(x) as a substep in their struggle to compute the derivative of cos(x). I don't think the suggestion is really that appropriate - either we know the derivative of cos(x) is sin(x) and we should use that fact directly, or we should do some other analysis on the limit that doesn't involve differentiation. I'll let op opine on whether they have learned the derivative of cos(x) yet

member 731016
Office_Shredder said:
I find it a bit weird to suggest someone compute the derivative of cos(x) as a substep in their struggle to compute the derivative of cos(x). I don't think the suggestion is really that appropriate - either we know the derivative of cos(x) is sin(x) and we should use that fact directly, or we should do some other analysis on the limit that doesn't involve differentiation. I'll let op opine on whether they have learned the derivative of cos(x) yet
This is true. Brain fart on my part. I guess I just looked at the last limit without reading the rest ...

To compute the derivative I would then apply either the derivative of ##e^{ix}##, which should be known to be ##i e^{ix}## or use the series definition of the cosine from which it should be clear that ##\cos(x) = 1 - x^2/2 + \mathcal O(x^4)##, which is sufficient to get the result.

member 731016
ChiralSuperfields said:
##\lim_{x \to 0^+} \frac{\cos x - 1}{x}## Howver, I'm not sure where to go from here.
Recall that ## 2\sin^2x = 1-\cos 2x ## and consider the behaviour of ##\frac{\sin t}{t}## as ##t\to 0##.

member 731016

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