Show that the function is not differentiable at the point z

[mathmari]

Hi! Could you help me with the following?
Let $$g: R \rightarrow R$$ a bounded function. There is a point $$z \epsilon R$$ for which the function $$h: R$$ \ $$\{z\} \rightarrow R$$, where $$h(x)=\frac{g(x)-g(z)}{x-z}$$ is not bounded. Show that the function $$g$$ is not differentiable at the point $$z$$.

My idea is the following:
Let the function $$g$$ be differentiable at the point $$z$$, so $$lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L$$. So there is a ε>0: $$|\frac{g(x)-g(z)}{x-z}-L|<ε$$.
$$|h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L|$$.
So the function $$h$$ is bounded. That cannot be true. So the function $$g$$ is not differentiable at the point $$z$$.

Is my idea right??

 

[Opalg]

Hi! Could you help me with the following?
Let $$g: R \rightarrow R$$ a bounded function. There is a point $$z \epsilon R$$ for which the function $$h: R$$ \ $$\{z\} \rightarrow R$$, where $$h(x)=\frac{g(x)-g(z)}{x-z}$$ is not bounded. Show that the function $$g$$ is not differentiable at the point $$z$$.

My idea is the following:
Let the function $$g$$ be differentiable at the point $$z$$, so $$lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L$$. So there is a ε>0: $$|\frac{g(x)-g(z)}{x-z}-L|<ε$$.
$$|h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L|$$.
So the function $$h$$ is bounded. That cannot be true. So the function $$g$$ is not differentiable at the point $$z$$.

Is my idea right??

Absolutely correct! (Sun)

 

[mathmari]

Great! (Clapping)

 

[mathmari]

Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?

 

[Opalg]

Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?

Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.

 

[mathmari]

Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.

So do I have to show both cases $ |x-z|< \delta$ and $ |x-z|\geq \delta$ ??