• Support PF! Buy your school textbooks, materials and every day products Here!

Limit Proof

  • Thread starter Math_Geek
  • Start date
23
0
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


2. Homework 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.

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

Answers and Replies

1,631
4
well, what you need to show that, [tex]\lim_{x\rightarrow a}\frac{1}{f(x)}=0[/tex], in epsilon delta we have: [tex]\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon [/tex].

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

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

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

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:
23
0
i understand but what is [tex]? at the top, I am not sure what that is.
thanks.
 
1,631
4
i understand but what is [tex]? at the top, I am not sure what that is.
thanks.
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:
1,631
4
that thing about shifting the bounds doesn't really make sens to me.
 
23
0
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:
1,631
4
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 [tex]\lim_{x\rightarrow a}\frac{1}{f(x)}=0[/tex], right?? using the fact that is in there. So using epsilon delta language, it actually means that we need to show that: [tex]\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon [/tex]
Right?

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

so our line at the top is fullfiled now, because we found that for every epsilon, and also for [tex]\epsilon=\frac{1}{M},\exists\delta(\epsilon)>0,[/tex] such that whenever, [tex]0<|x-a|<\delta=>|\frac{1}{f(x)}|<\frac{1}{M}=\epsilon [/tex] this is all we needed to show.
 
23
0
got it thanks
 
1,631
4
23
0
yes it is!
 

Related Threads for: Limit Proof

  • Last Post
Replies
3
Views
945
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
954
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
543
  • Last Post
Replies
4
Views
394
  • Last Post
Replies
1
Views
853
Top