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?
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 $a=0$.