Does g(x) Approach Infinity Given f(x)'s Limit and Inequality Conditions?

[Math_Geek]

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

Homework Equations

using delta epsilon

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

 

[sutupidmath]

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

Homework Equations

using delta epsilon

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:

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


we know that: $$g(x)\geq f(x), \forall xE R$$, and we also know that

$$\lim_{x\rightarrow a}f(x)=\infty$$

in $$\epsilon,\delta$$ language this actually means

For any M>0, $$\exists\delta>0$$ such that whenever $$0<|x-a|<\delta$$, (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 $$f(x)>M$$. But from here since

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

Following this line of reasoning we have that

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

$$\lim_{x\rightarrow a}g(x)=\infty$$


Hope this helps..

 

[sutupidmath]

You, do not have to use epsilon in this case, since both these functions, g, f, obviously diverge to infinity. The expression $$\lim_{x\rightarrow a}g(x)=\infty$$
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.

 

[sutupidmath]

1. we have that g(x)<e.

YOu got it right up to this point.

 

[tiny-tim]

… 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!

Well … actually … you've got it right!

(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 … )

Now write it out in proper English!​