Limits at Infinity proof

Question

The problem is prove the limit using definition 6, $$\lim_{x\rightarrow-3} \frac{1}{(x+3)^4} = \infty$$

The book gives definition 6 as:

Let $f$ be a function defined on some open interval that contains $a$, except possibly at $a$ itself. Then $\lim_{x\rightarrow a} f(x) = \infty$ means that for every positive number $E$ there is a positive number $\delta$ such that $0 <|x-a| < \delta$ then $f(x) > E$.

Can you please explain it step by step all the way to the answer? Thank you.

You do not understand the definition or the use of the definition? — Yes, Sep 20 '15 at 03:56
I don't understand the definition or the use of the definition. I just started learning calculus last month. — Anonymous, Sep 20 '15 at 03:59

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

The definition yields a rough plan of what you need to do to show $\lim_{x\to a} f(x)=\infty$:

Fix an arbitrary $E>0$.
Define appropiate $\delta >0$, independent of $x$ (dependency on $E$ and $a$ is fine).
You can assume that for $x$ you have $|x-a|<\delta$.
With the knowledge of 1.-3. you now need to show that $f(x)>E$ (for every $x$ as in 3.).

Action point 2. is the key here and usually that's where you need to get creative. The problem is to pick an appropiate $\delta$ which just means you need to find a $\delta$ so that step 4. can be carried out. But how is it possible to know beforehand (in step 2 out of 4) what $\delta$ to pick so that it works out in the end? Short answer: In most cases this is just not possible. Standard advice is to "work backwards", which basically means you look at what you want to show (4.) and try to get to what you have been given (1.-3.). (Further reading.)

The use of the definition in your case $f(x)=(x+3)^{-4}$ is not as problematic as described above since $\delta$ is found rather quickly (see below). You might want to try this alone first by combining what I wrote above and the hint in the answer by AjmalW.

1. Fix arbitrary $E>0$.
2. Define $\delta:=1/\sqrt[4]{E}$.
3. Assume $|x+3|<\delta$.
4. Then we have
$$f(x)=\frac{1}{(x+3)^4}\overset{3.}{>}\frac{1}{\delta^4}\overset{2.}{=}E,$$
so $f(x)>E$ as desired. Since $E>0$ was arbitrary (1.) we just showed $\lim_{x\to -3} f(x)=\infty$ according to the definition.

Finding $\pmb\delta$ might be approached like this:

We need to show $f(x)=\frac{1}{(x+3)^4}>E$ and we are given control over $|x+3|$ (in the end we can pick any $\delta$ such that $|x+3|<\delta$). So we transform $\frac{1}{(x+3)^4}>E$ into something else of which we know more:
$$\frac{1}{(x+3)^4}>E \Longleftrightarrow \frac{1}{E}>|x+3|^4\Longleftrightarrow \frac{1}{\sqrt[4]{E}}>|x+3|$$

So if we can make the last inequality true, we are finished. But looking at 3., we in fact can assume that inequality is true if we only pick $\delta$ to be $1/\sqrt[4]{E}$ in step 2.

Limits at Infinity proof

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