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- Homework Statement
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- Relevant Equations
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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.

Does someone please know?

Thanks!

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.

Does someone please know?

Thanks!