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Let f be a real valued function satisfying

$|f (x) −f (a)| ≤ C|x−a|^γ$,

for some γ > 0 and C >0.

(a) If γ = 1, show that f is continuous at a;

(b) If γ > 1, show that f is differentiable at a.

Now i solved (a) by taking a function g(x)=x which is continuous at x=a and using it to show that for every C$\epsilon$*=$\epsilon$>0 there exists a $\delta$>0 such that $|f (x) −f (a)| ≤ C|x−a|<\epsilon$ whenever $|x-a|<\delta$.

Now for (b) when $\gamma$ > 1,

we get $|\frac{f(x)-f(a)}{x-a}|$< C$|x-a|^{\gamma-1}$, which gives that $\displaystyle \lim_{x \rightarrow a} \left(\frac{f(x)-f(a)}{x-a}\right)$=0 by sandwich theorem.

We define the first derivative for a function $f(x)$ at x=a to be

$\displaystyle \lim_{x \rightarrow a} \left(\frac{f(x)-f(a)}{x-a}\right)$

if it exists.

My question is if the above limit goes to zero then does it mean that f(x) is not differentiable at x=a or has an extremum at x=a?

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  • $\begingroup$ This limit goes to $0$ is the definition of $f'(a)=0$ which means that $f$ has a local extremum in $a$ $\endgroup$
    – Vincent
    Apr 26, 2016 at 21:18

2 Answers 2

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I'll assume $f$ is defined in a neighborhood of $a$. Note that you cannot use a particular $f$ for proving the result.

For the case $\gamma>0$ (in particular $\gamma=1$), the squeeze theorem suffices: from $\lim_{x\to a}|x-a|^\gamma=0$ it follows that $$ \lim_{x\to0}|f(x)-f(a)|=0 $$ so also $\lim_{x\to0}(f(x)-f(a))=0$ and, finally, $$ \lim_{x\to0}f(x)=f(a) $$

Suppose now that $\gamma>1$. Then, for $x\ne a$, $$ 0\le\left|\frac{f(x)-f(a)}{x-a}\right|\le C|x-a|^{\gamma-1} $$ Again, the squeeze theorem yields $$ \lim_{x\to a}\frac{f(x)-f(a)}{x-a}=0 $$ so $f$ is differentiable at $a$ and $f'(a)=0$.

The fact that the derivative is $0$ means that $a$ can be an extremal point, but not necessarily: consider $f(x)=x^3$ and $a=0$.

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$$\lim_{x \to a}\frac{f(x) - f(a)}{x - a} = 0$$

means that $f$ is differentiable at $a$ and $f'(a) = 0$ (by definition).

If $f'(a) = 0$, then $f$ doesn't necessarily have a (global or local) extremum at $x = a$. For instance, take $f(x) = x^3$ and $a = 0$.

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  • $\begingroup$ So is it sufficient to show that the above limit goes to 0 to prove question (b) i.e. f(x) is differentiable at x=a? @AhmedHussein $\endgroup$
    – user666
    Apr 26, 2016 at 21:23
  • $\begingroup$ @user111986 yes. What else can you possibly do? This is the definition of differentiability $\endgroup$
    – user258700
    Apr 26, 2016 at 21:26
  • $\begingroup$ Alright. Thanks. @AhmedHussein $\endgroup$
    – user666
    Apr 26, 2016 at 21:28

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