limits of functions

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February 15th 2009, 01:55 PM
claire511

limits of functions

Im struggling with a couple of questions... any help would be appreciated. You have to answer true or false and if true give a proof, if false give a counterexample.

(a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.

I think this is true but i have no idea how to prove it, I wouldn't even know where to start!

(b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.

I think this must be false because of the subtle difference in the question - is it false because the limit of the derivative function might not actually exist? This seems to make sense to me as the only way it could be false, although I cannot think of a counterexample.

thanks
February 15th 2009, 03:40 PM
Reckoner

Quote:

Originally Posted by claire511

(a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.

(b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.

For the second, here is a counterexample: Consider

$f(x) = \left\{\begin{array}{ll} \frac1x, & x\text{ is rational}\\ 0, & x\text{ is irrational} \end{array}\right.\text.$

Now, $\lim_{x\to\infty}f(x)=0,$ and is thus finite, which is easy to show by the squeeze/sandwich theorem, using $x\mapsto\frac1x$ and $x\mapsto0\text.$

But, $f(x)$ is nowhere continuous. To show this, choose any $c\in\mathbb R\text.$

First, if $c$ is rational, then $f(c) = \frac1c$ . Take $\varepsilon=\frac1{2c}\text.$ Now, no matter how small you choose $\delta>0$ to be, there is always going to be an irrational number $d$ within $\delta$ units of $c,$ in which case $\left\lvert f(d)-\frac1c\right\rvert=\left\lvert0-\frac1c\right\rvert=\frac1c>\frac1{2c}=\varepsilon \text.$ Thus, $\lim_{x\to c}f(x)\neq f(c)$ and $f$ is not continuous at $c\text.$ The case for irrational $c$ should be similar.

And of course, if $f$ is everywhere discontinuous, then at any given value its derivative is undefined.

Does that help any? I could probably get you started on (a), but you'll have to wait a bit.
February 15th 2009, 08:39 PM
CaptainBlack

Quote:

Originally Posted by claire511

(b) If lim(f(x)) as x tends to infinity is finite, the lim(f'(x))=0.

Consider:

$ f(x)=\frac{\sin(x^2)}{x} $

CB
February 15th 2009, 08:51 PM
CaptainBlack

Quote:

Originally Posted by claire511

(a) If lim(f(x)) as x tends to infinity is finite and limf'(x)=b then b=0.

If $b>0$ then for $x$ large enough $f'(x)>\epsilon>0$ , and so for $x$ large enough and all $y>0$ (by a simple corollary to the mean value theorem):

$f(x+y)>f(x)+\epsilon y$

hence the limit of this as $y \to \infty$ cannot be finite, which contradicts the assumption that $\lim_{x \to \infty}f(x)$ is finite.

A similar argument applies when $b<0$ .

Now you will need to fill in the detail to complete this.

CB
February 16th 2009, 10:53 AM
claire511

Quote:

Originally Posted by CaptainBlack

If $b>0$ then for $x$ large enough $f'(x)>\epsilon>0$ , and so for $x$ large enough and all $y>0$ (by a simple corollary to the mean value theorem):

$f(x+y)>f(x)+\epsilon y$

hence the limit of this as $y \to \infty$ cannot be finite, which contradicts the assumption that $\lim_{x \to \infty}f(x)$ is finite.

A similar argument applies when $b<0$ .

Now you will need to fill in the detail to complete this.

CB

Thank you so much for replying. I understand by corollary to MVT if y positive f(x+y)>f(x) but why is there a plus epsilon times y bit on the end :S?? then how does letting y tend to infinity show that the limit of x tending to infinity of f(x) is not finite?
February 16th 2009, 11:35 AM
CaptainBlack

Quote:

Originally Posted by claire511

Thank you so much for replying. I understand by corollary to MVT if y positive f(x+y)>f(x) but why is there a plus epsilon times y bit on the end :S?? then how does letting y tend to infinity show that the limit of x tending to infinity of f(x) is not finite?

If $b>0$ then for $x$ large enough $f'(x)$ never strays far from $b$ , and so $f'(x)$ is strictly greater than some positive constant (which we choose to call $\epsilon$ ) Then if we know that between $x$ and $y$ (with $y>0$ ) the derivetive is always greater than $\epsilon>0$ then: $f(x+y)-f(x)>\epsilon y$

Now we are considering $x$ as a constant, so:

$\lim_{u \to \infty} f(u)=\lim_{y \to \infty} f(x+y)$

but we see from earlier that $\lim_{y \to \infty} f(x+y) =\infty$

CB