Proving the Limit of 1/f(x) When f(x) Goes to Infinity

[Math_Geek]

1. Homework Statement [/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

Homework Equations

Need to show with a delta-epsilon proof

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.

Am I close? Please help a girl in distress! lol
Michelle

 

[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!

 

[Math_Geek]

i understand but what is $$? at the top, I am not sure what that is.<br /> thanks.$$

 

[sutupidmath]

i understand but what is $$? at the top, I am not sure what that is.<br /> thanks.$$

[tex] <br /> 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 $$anywhere neither at the top nor at the bottom.$$[/tex]

 

[sutupidmath]

that thing about shifting the bounds doesn't really make sens to me.

 

[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.

 

[sutupidmath]

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?

We, like i wrote before on my other post: We need to show that $$\lim_{x\rightarrow a}\frac{1}{f(x)}=0$$, 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.

 

[Math_Geek]

got it thanks

 

[sutupidmath]

got it

Good job!

thanks

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

 

[Math_Geek]

yes it is!