1. Limit and Derivative proof

Let $f0,1) \rightarrow \mathbb {R} " alt="f0,1) \rightarrow \mathbb {R} " /> be differentiable and let $f'(x)$ be Bounded. Show that $\lim _{x \rightarrow 0^+ } f(x)$ exist.

Idea so far.

So if I define a sequence that converges to a point at 0, f would be able to able to map that into the co-domain, would that be the right start?

Thanks.

Let $f0,1) \rightarrow \mathbb {R} " alt="f0,1) \rightarrow \mathbb {R} " /> be differentiable and let $f'(x)$ be founded. [I'll assume that should be "bounded".] Show that $\lim _{x \rightarrow 0^+ } f(x)$ exist.

Idea so far.

So if I define a sequence that converges to a point at 0, f would be able to able to map that into the co-domain, would that be the right start?
Yes, that is a good way to start. Now prove that the sequence $f(x_n)$ is Cauchy, and hence has a limit. Then you'll need to check that this limit is the same for all sequences $x_n\to0$.

(A quicker way to say the same thing is that a function with a bounded derivative is uniformly continuous; and if a function is uniformly continuous on a set A then it extends continuously to the closure of A.)

3. So in a differentiable function f, with f'(x) bounded, any sequence would be cauchy if it is mapped by f?

So in a differentiable function f, with f'(x) bounded, any sequence would be cauchy if it is mapped by f?
Yes, by the mean value theorem. If $|f'(x)|\leqslant M$ for all x, then $f(y)-f(x)=f'(\xi)(y-x)$ for some ξ between x and y, and so $|f(y)-f(x)|\leqslant M|y-x|$. If $(x_n)$ is Cauchy then $|f(x_m)-f(x_n)|\leqslant M|x_m-x_n|$, from which it follows that $(f(x_n))$ is also Cauchy.

5. Okay, so I'm picking up from after I know that $f(x_n)$ converges.

Now, suppose that $f(x_n) \rightarrow L$.

Since f is differetiable on (0,1), it is also continuous on (0,1), implies that $f(x_n) \rightarrow f(0)$ since $x_n \rightarrow 0$.

So is this enough to show that $\lim _{x \rightarrow 0^+} f(x)$ exist and it is also equals to $f(0)$?

A function with a bounded derivative is uniformly continuous.
Now go back to the last question I helped you with.
I think that was a lemma for this question.

7. let $\epsilon > 0$ be given, and find $\delta > 0$, and suppose that the sequence $\{ x_n \}$ converges to 0. Then $\exists N \in \mathbb {N}$ such that whenever $n \geq N$ implies that $d(x_n,0)< \delta$

Since f is uniformly continuous, we have $p(f(x_n),f(0)) < \epsilon$

So for all sequences that converges to 0, the sequences after being mapped by the function f would converge to f(0).

I hope this is right, suicide or not.

Thanks!!!