Showing a limit exists using differentiability

[B3NR4Y]

Homework Statement

Assume f:(a,b)→ℝ is differentiable on (a,b) and that |f'(x)| < 1 for all x in (a,b). Let a_n
be a sequence in (a,b) so that a_n→a. Show that the limit as n goes to infinity of f(a_n) exists.

Homework Equations

We've learned about the mean value theorem, and all of that fun stuff.

The Attempt at a Solution

I don't really know where to start so I brainstormed a couple of things I noticed[/B]
I know that since |f'(x)| is always less than one, any sequence of points will be bounded. Since they are bounded they are cauchy.
I also know that
$lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$ exists for all x. I assume this should also mean that $lim_{n \rightarrow \infty} \frac{f(a_n)-f(a)}{a_n - a}$ exists. I think with these two facts I can construct a proof but I don't know if either of the two are correct

 

[Samy_A]

Homework Statement

Assume f:(a,b)→ℝ is differentiable on (a,b) and that |f'(x)| < 1 for all x in (a,b). Let a_n
be a sequence in (a,b) so that a_n→a. Show that the limit as n goes to infinity of f(a_n) exists.

Homework Equations

We've learned about the mean value theorem, and all of that fun stuff.

The Attempt at a Solution

I don't really know where to start so I brainstormed a couple of things I noticed[/B]
I know that since |f'(x)| is always less than one, any sequence of points will be bounded. Since they are bounded they are cauchy.
I also know that
$lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$ exists for all x. I assume this should also mean that $lim_{n \rightarrow \infty} \frac{f(a_n)-f(a)}{a_n - a}$ exists. I think with these two facts I can construct a proof but I don't know if either of the two are correct

Hint:

Take two element of the sequence, $a_n$ and $a_m$, and apply the mean value theorem for these two values.
That should give you an interesting upper bound for $|f(a_n)-f(a_m)|$.

It is not true that a bounded sequence is Cauchy.
But, a convergent sequence is a Cauchy sequence, and in ℝ a Cauchy sequence has a limit. This, together with the upper bound for $|f(a_n)-f(a_m)|$ should lead to a solution.

 

[B3NR4Y]

Ok so applying the mean value theorem, there is an x in (a_m,a_n) so that $\frac{f(a_m) - f(a_n)}{a_m - a_n} = f'(x)$ taking the absolute value of both sides, $|f(a_m) - f(a_n)| < |a_m - a_n|$. Since a_n is convergent, |a_m - a_n| is always less than some epsilon greater than zero (because it is a convergent sequence in R and therefore Cauchy), and so |f(a_m) - f(a_n)| is also always less than that epsilon. This is the definition of a cauchy sequence, so the sequence f(a_n) must have a limit.

Is this the right path?

 

[Samy_A]

Ok so applying the mean value theorem, there is an x in (a_m,a_n) so that $\frac{f(a_m) - f(a_n)}{a_m - a_n} = f'(x)$ taking the absolute value of both sides, $|f(a_m) - f(a_n)| < |a_m - a_n|$. Since a_n is convergent, |a_m - a_n| is always less than some epsilon greater than zero (because it is a convergent sequence in R and therefore Cauchy), and so |f(a_m) - f(a_n)| is also always less than that epsilon. This is the definition of a cauchy sequence, so the sequence f(a_n) must have a limit.

Is this the right path?

Yes.
Essentially you have that $(f(a_n))_n$ is a Cauchy sequence, therefore convergent.

Maybe a remark: where you say "is always less than some epsilon", you should add "for n and m sufficiently large". But that is probably what you implicitely meant.