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If $f$ is an even function such that $\lim_{h\to 0^+} \frac{f(h)-f(0)}{h}$ has some finite non zero value, prove that $f(x)$ is continuous but not differentiable.

$$f’(0)=\lim_{h\to 0^+} \frac{ f(h)-f(0)}{h} =k$$ And $$f’(0)=\lim_{h\to 0^-} \frac{ f(0)-f(-h)}{h}=-k$$

So function is not differentiable at $x=0$

But how do I prove that it is continuous at $x=0$?

Kyan Cheung
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Aditya
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  • If $$\lim_{h\to 0} \frac{ f(h)-f(0)}{h} =k$$ it necessarily follows that $$\lim_{h\to 0} (f(h)-f(0)) =0$$ so $f$ is continuous in $0$. In general it's not continuous on the whole domain. – Gono Sep 13 '20 at 11:37
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    The wording is quite bad. If $\lim_{h\to 0} \frac{f(h) - f(0)}{h}$ exists, then by definition $f$ is differentiable at $0$. Probably you mean $h\to 0^+$. – Arctic Char Sep 13 '20 at 11:44

1 Answers1

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Recall that we define a function $f$ to be continuous at point $c$ iff $$\lim_{x\to c}f(x)=f(c)$$ We need to show $\lim_{h\to0}f(h)=f(0)$.

For the right hand limit:

$$\lim_{h\to0^+}f(h)=f(0)+\lim_{h\to0^+}[f(h)-f(0)]=f(0)+\lim_{x\to0^+}h\left[\frac{f(h)-f(0)}{h}\right]=f(0)+0\cdot k=f(0)$$ For the left hand limit:

$$\lim_{h\to0^-}f(h)=f(0)+\lim_{h\to0^-}[f(h)-f(0)]=f(0)+\lim_{x\to0^-}h\left[\frac{f(h)-f(0)}{h}\right]=f(0)+\lim_{x\to0^-}\left[-h\frac{f(0)-f(h)}{h}\right]=f(0)+\lim_{x\to0^-}\left[-h\frac{f(0)-f(-h)}{h}\right]=f(0)-0\cdot k=f(0)$$ Combining the two

$$\lim_{x\to0}f(h)=f(0)$$

And thus $f$ is continuous at $x=0$.

Kyan Cheung
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