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One more for the road!

  • Thread starter Math_Geek
  • Start date
23
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1. Homework Statement
Prove: If the Limit as x goes to a of f(x)=infinity and g(x)>or equal to f(x) for all x in the reals, then limit as x goes to a of g(x) is inf


2. Homework Equations
using delta epsilon


3. The Attempt at a Solution
for all e>0 there exist a delta>0 s.t g(x)<e now using lim of f(x) goes to infinity means there is an M>0, there exists a delta>0 we get f(x)>M so for 0<|x-a|<delta we have f(x)>M=e so therefore since g(x) is greater than f(x), we have that g(x)<e. This is probably wrong, so any help is greatly appreciated
 

Answers and Replies

1,631
4
1. Homework Statement
Prove: If the Limit as x goes to a of f(x)=infinity and g(x)>or equal to f(x) for all x in the reals, then limit as x goes to a of g(x) is inf


2. Homework Equations
using delta epsilon


3. The Attempt at a Solution
for all e>0 there exist a delta>0 s.t g(x)<e now using lim of f(x) goes to infinity means there is an M>0, there exists a delta>0 we get f(x)>M so for 0<|x-a|<delta we have f(x)>M=e so therefore since g(x) is greater than f(x), we have that g(x)<e. This is probably wrong, so any help is greatly appreciated

Ok, this is simmilar to the previous one.
what we need to show is that:

[tex]\lim_{x\rightarrow a}g(x)=\infty[/tex], in epsilon delta language, this means :
that for any M>0, [tex]\exists\delta>0[/tex] such that whenever [tex]0<|x-a|<\delta[/tex] we have [tex] g(x)>M[/tex]--------------(*)


we know that: [tex]g(x)\geq f(x), \forall xE R[/tex], and we also know that

[tex]\lim_{x\rightarrow a}f(x)=\infty[/tex]

in [tex]\epsilon,\delta[/tex] language this actually means

For any M>0, [tex]\exists\delta>0[/tex] such that whenever [tex]0<|x-a|<\delta[/tex], (lets supposte that this M is the same as that used in (*). Or if we wish not so, we can simply chose their maximum. ) we have [tex]f(x)>M[/tex]. But from here since

[tex]g(x)\geq f(x), \forall xE R[/tex], it means that [tex]g(x)\geq f(x)[/tex] also for x-s within the interval [tex](a-\delta,a+\delta)[/tex].

Following this line of reasoning we have that

For any M>0,[tex]\exists\delta>0[/tex], such that whenever [tex]0<|x-a|<\delta[/tex], we have [tex]g(x)\geq f(x)>M[/tex], which actually means nothing else but that:

[tex]\lim_{x\rightarrow a}g(x)=\infty[/tex]


Hope this helps..
 
1,631
4
You, do not have to use epsilon in this case, since both these functions, g, f, obviously diverge to infinity. The expression [tex]\lim_{x\rightarrow a}g(x)=\infty[/tex]
simply tells us that the function g increases without bound, and nothing else. It does not say that g(x) equal infinity, since that does not make sense.
 
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tiny-tim
Science Advisor
Homework Helper
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… by George, she's got it … !

For all e>0 there exist a delta>0 s.t g(x)<e now using lim of f(x) goes to infinity means there is an M>0, there exists a delta>0 we get f(x)>M so for 0<|x-a|<delta we have f(x)>M=e so therefore since g(x) is greater than f(x), we have that g(x)<e. This is probably wrong, so any help is greatly appreciated
Hi Michelle! :smile:

Well … actually … you've got it right! :rofl:

(except, of course, that you should have written "g(x) > e" in the first and last lines! and M = e, so you didn't need both … :redface:)

Now write it out in proper English!
 

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