# Limit Proof

1. Mar 24, 2008

### Math_Geek

1. The problem statement, all variables and given/known data[/bProve
Prove: If the limit as x goes to a of f(x)=infinity, then lim as x goes to a of 1/(f(x) =0

2. Relevant equations
Need to show with a delta-epsilon proof

3. The attempt at a solution

using the definition, lim as x goes to a f(x)=infinity means that for any M>0 there exists an delta>0, where a<x<a+delta implies that f(x)>M. So using this def, I know there is M>0, there exists a delta (not sure what yet) so that a<x<a+delta and taking 1/f(x) shifts the bounds a+delta<x<a and then M would be less than or equal to 0 therefore the lim as x goes to a of 1/f(x)=0.

Michelle

2. Mar 24, 2008

### sutupidmath

well, what you need to show that, $$\lim_{x\rightarrow a}\frac{1}{f(x)}=0$$, in epsilon delta we have: $$\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon$$.

using the fact that $$\lim_{x\rightarrow a}f(x)=\infty$$, as u stated, with a little omission, means that: For any M>0, $$\exists\delta>0$$ such that whenever

$$0<|x-a|<\delta$$ we get f(x)>M. NOw using this fact here we get that whenever

$$0<|x-a|<\delta$$ we have $$\frac{1}{f(x)}<\frac{1}{M}=\epsilon=>|\frac{1}{f(x)}|<\epsilon=\frac{1}{M}$$

Now you only need to put everything together, because it is a little messy.

I hope this helps. But feel free to ask again, don't put too much stress on yourself!

Last edited: Mar 24, 2008
3. Mar 24, 2008

i understand but what is $$? at the top, I am not sure what that is. thanks. 4. Mar 24, 2008 ### sutupidmath I don't know what are u talking about? I just used latex to write those math symbols, probbably your browser, or sth, is not being able to generate those symbols, because i do not see [tex] anywhere neither at the top nor at the bottom. Last edited: Mar 25, 2008 5. Mar 25, 2008 ### sutupidmath that thing about shifting the bounds doesn't really make sens to me. 6. Mar 25, 2008 ### Math_Geek it is something our teacher showed us how to do, but it never seems to work for me. One more question, we (meaning you) showed the limit is equal to zero becasue 1/M <epsilon? I get confused because we let m>0. so how can the limit be =0. Last edited: Mar 25, 2008 7. Mar 25, 2008 ### sutupidmath We, like i wrote before on my other post: We need to show that [tex]\lim_{x\rightarrow a}\frac{1}{f(x)}=0$$, right?? using the fact that is in there. So using epsilon delta language, it actually means that we need to show that: $$\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon$$
Right?

Now from:
$$\lim_{x\rightarrow a}f(x)=\infty$$
we know that for every M>0 ,$$\exists\delta>0$$ such that for every $$xE(a-\delta,a+\delta),$$ with the possible exception when x=a, we have $$f(x)>M$$, but also for that delta and for that interval we also have $$\frac{1}{f(x)}<\frac{1}{M}=\epsilon=>|\frac{1}{f(x )}|<\epsilon=\frac{1}{M}$$

so our line at the top is fullfiled now, because we found that for every epsilon, and also for $$\epsilon=\frac{1}{M},\exists\delta(\epsilon)>0,$$ such that whenever, $$0<|x-a|<\delta=>|\frac{1}{f(x)}|<\frac{1}{M}=\epsilon$$ this is all we needed to show.

8. Mar 25, 2008

### Math_Geek

got it thanks

9. Mar 25, 2008

### sutupidmath

Good job!

No, problem.
I love the idea behind this forum. It is blessed.

10. Mar 25, 2008

yes it is!