A limit proof with a sequence

Say I want to prove a limit: $$ \underset{x\ \rightarrow \ -1}{\lim} \frac{1}{(x + 1)^4} = \infty $$

Can I prove it by using the sqeuence: $ a_n = -1 + \frac{1}{n} $: $$ \underset{n\ \rightarrow \ \infty}{\lim} \frac{1}{(\frac{1}{n})^4} = \underset{n\ \rightarrow \ \infty}{\lim} n^4 = \infty$$

I just don't remember if that is okay to use this method when proving or only when disprooving a limit. Thanks!

Solution 1:

The idea behind it is good but incomplete.

In fact you proved it goes to infinity but only for one particular sequence, you need to prove it unconditionally.

It is not too difficult to remedy to this, since $x\to -1$ this means $x=-1+u$ with $u\to 0$.

See how we choose a general $u$ instead of $\dfrac 1n$.

Another problem of your method is that $\frac 1n$ is always a positive quantity, but nothing prevent $u$ to have a random sign.

However since we consider $(x+1)^4$ it doesn't matters here because the power $4$ eliminates the sign issue.

Therefore WLOG we can consider $u=\dfrac 1v$ with $v\to +\infty$ (but would have it been $(x+1)^3$ you would have to consider the sign of $v$).

Finally we get $\dfrac 1{(x+1)^4}=\dfrac 1{u^4}=v^4\to+\infty$.

A complement based on Lazy's comment.

What is lacking in your method is showing that the function has a limit. If it has a limit then by substituting $-1+\frac 1n$ you proved that this limit is necessarily $+\infty$.

This can effectively be achieved by showing the function is increasing toward $-1$.

Solution 2:

(Don't forget also that you need to search also for $x<-1$ and $x>1$)

In your situation, you need that :

$\forall (x_n)_{n\in\mathbb N}\in\mathbb ]-1, +\infty[^\mathbb N, $ such that $x\to -1$, $f(x_n)\to +\infty$

You might not understand why the need of the $\forall$ here. Basically, imagine another function like $\sin(x)$.

If I choose $((-1)^n\pi + 2\pi n)_{n\in\mathbb N}$, when $n\to +\infty$, you get $\sin((-1)^n\pi + 2\pi n) \to 0$. However, $\sin(x)$ diverges when $x\to+\infty$ !

Another one, $f(x)=e^{1/(x+1)}\cos\left(\frac{1}{x+1}\right)$ for $x>-1$.

We have $f\left(\frac{1}{2\pi n}-1\right) = e^{2\pi n}$ goes to $+\infty$. Yet $f\left(\frac{1}{2\pi n + \pi}-1\right) = -e^{2\pi n + \pi}$ goes to $-\infty$.

So when $x\to-1$, $f(x)$ just diverges (not $+$ or $-\infty$).

So even if you have a sequence who does work, it is still possible that you don't get the real answer.

P.S : Your sequence give you just one information. It just specifically goes to $+\infty$ or diverges.