Originally Posted by

**nein12** Oops... my bad.

If lim x->a g(x) = infinity and g(x) < or equal to f(x) for x -> a, then limx-> a f(x) = infinity

Prove, using the formal definition of limits, each of the statement is true.

So... that's the question. Honestly, I never had to answer this kind of questions. It's usually applications format or equation format.

Another one is limx->infinity f(x) = -infinuty and c > 0, then limx ->infinity c f(x) = - infinity

Not sure which rules to use, but I can reason... that f(x) is a negative infinity... times bu any constant positive number is still negative infinity. I am not sure if there are any "specific rules".

So we wish to show that if $\displaystyle \lim_{x\to{a}}g(x)=\infty$ and $\displaystyle g(x)\leq{f(x)}$ *as ? $\displaystyle x\to{a}$ ?, *then $\displaystyle \lim_{x\to{a}}f(x)=\infty$

Well, for infinite limits the definition states:

Let g(x) be a function that is defined on an interval containing a, except possibly at a. Then we say that

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

if for every number M>0, there is some number $\displaystyle \delta>0$ such that

$\displaystyle g(x)>M$ whenever $\displaystyle 0<\mid{x-a}\mid<\delta$

So then the proof is straight forward

if $\displaystyle g(x)>M$ for every x arbitrarily close to a, then $\displaystyle f(x)>g(x)$ for every value of x arbitrarily close to a $\displaystyle \Rightarrow{f(x)>M}$

Therefore

$\displaystyle f(x)>M$ whenever $\displaystyle 0<\mid{x-a}\mid<\delta$