# I have to proof that $\lim_{x \to \c} frac{1}{f(x) = 0$

1. Dec 8, 2012

### kasperrepsak

1. The problem statement, all variables and given/known data
Given is: f is a function that maps D onto the Real numbers, and c is within D and it is a limit point, and f(x) =/= 0 for all x in D, and $\lim_{x \to c} f(x) = \infty$
I have to proof that:
$\lim_{x \to c} \frac{1}{f(x)} = 0$

3. The attempt at a solution

This means that according to the definition I have to proof that $\forall \epsilon \ \exists \delta$ so that $\forall x \in D \ with \ 0 < |x-c|< \delta \ \ : |\frac{1}{f(x)} - 0|< \epsilon$.

Im not sure how to go on from here. Or do i have to do something else?

2. Dec 8, 2012

### jbunniii

Hint:
$$\left|\frac{1}{f(x)} - 0\right| < \epsilon$$
is equivalent to
$$|f(x)| > \frac{1}{\epsilon}$$

3. Dec 8, 2012

### kasperrepsak

Would it then be enough to say that because f diverges to infinity as x approaches c, this by definition means that there must be a $\delta$ for which if $0<|x-c|<\delta$ the latter inequality that you wrote holds, which is equivalent to the former, quod erat demonstrandum?

Last edited: Dec 8, 2012
4. Dec 8, 2012

### SammyS

Staff Emeritus
What you have to prove is what you have in the following statement .
What will help you get there is to state what is meant by:

$\displaystyle \lim_{x \to c} f(x) = \infty\ .$

5. Dec 9, 2012

### kasperrepsak

Well the weird thing is that in our book we hv only defined what is meant by a sequence diverging to infinity at a point c. This assignment is part of the chapter on continuity where weve only defined what is meant by a limit of f as x goes to infinity. But wut i wrote should b ok since that follows from the definition right?

6. Dec 9, 2012

### SammyS

Staff Emeritus
What you wrote is what needs to be proved.

What you need, among other things, to get there is a definition of what it means for $\displaystyle \lim_{x \to c} f(x) = \infty\ .$

You may be able to deduce this from
what is meant by a sequence diverging to infinity at a point c.​
and from
what it means for the limit of a function, f(x), to converge as x → c . ​

Basically, what it means for $\displaystyle \lim_{x \to c} f(x) = \infty\,,$ is that given any M > 0, (usually M is a large number) there exists a δ > 0 such that for any x satisfying 0<|x-c|<δ, you have that f(x) > M .

7. Dec 9, 2012

### kasperrepsak

No but i ment what i wrote in reply to the firs post. There must be a delta greater then 1/epsilon

8. Dec 9, 2012

### SammyS

Staff Emeritus
That works, if you spell out why it works.

Given an ε > 0, then 1/ε > 0 . Since $\displaystyle \ \lim_{x \to c} f(x) = \infty\ \$ there exists δ > 0, such that ... f(x)>1/ε ...