Proof check $f(x)=\frac{2x}{x-4}$ diverges as $x\to4$ using $\epsilon-\delta$

Question

My nephew asked me to consider $\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$ for a problem set he was doing.

I explained that it diverged by substituting $x=4+\delta$ and letting $\delta\to 0$.

I thought I'd try to do it formally using $\epsilon-\delta$ definition of divergence for practice as I haven't done one in years and I find them difficult.

Could someone please check the following as there's likely an error?

I used this definition: $\forall M>0, \exists\delta >0:|x-a|<\delta\implies f(x)>M$

For the problem at hand:

$\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$

We require:

$\forall M>0, \exists\delta >0:|x-4|<\delta\implies \frac{2x}{x-4}>M$

Some scratchwork is necessary

1st manipulate $\frac{2x}{x-4}>M$ as follows

$\left|\frac{2x}{x-4}\right|>|M|\implies \frac{|2x|}{|x-4|}>M$

Note that we may assume that $\delta\leq3\implies 1\leq x \leq7$

So $|2x|\geq2$

$\frac{|2x|}{|x-4|}\geq\frac{2}{|x-4|}>M$

So we may choose $\delta=\frac{2}{M}$

Proof: $\forall M>0 \exists \delta=\frac{2}{M}:|x-4|<\delta=\frac{2}{M}\implies \frac{2x}{x-4}>M$

Thank you

Answer 1

It is nearly fine. The only problems that I see are:

1. The final value for $\delta$ should be $\min\left\{\frac2M,3\right\}$, not just $\frac2M$.
2. What that actually proves is that $\lim_{x\to4^+}\frac{2x}{x-4}=\infty$. That is, what you actually proved is that, for any $M>0$, if you take the $\delta$ from above, then$$4<x<4+\delta\implies\frac{2x}{x-4}>M$$and that is enough to prove that that limit doesn't exist (in $\Bbb R$). But $\lim_{x\to4^-}\frac{2x}{x-4}=-\infty$.

Answer 2

Your definition of divergence is wrong, so I didn't read any further.

To prove: $\lim_{x \to 4} \frac{2x}{x-4}$ does not exist.

In order to get the correct epsilon-delta definition of divergence, you must specify the correct epsilon-delta definition of convergence, and then negate the definition.

The epsilon-delta definition of convergence is:

There exists a limit $L$ such that:
$\forall \epsilon > 0 ~~\exists ~\delta > 0$ such that
when $0 < |x - 4| < \delta$, then $|f(x) - L| < \epsilon,$
where $f(x) = \frac{2x}{x-4}$.

In order to prove divergence, you must therefore prove that:

For any proposed limit $L$,
a specific $\epsilon > 0$ can be found such that
regardless of how small you make $\delta$, where $0 < \delta$,
you will be able to find at least one $x$ such that both $0 < |x - 4| < \delta~$ and $~|f(x) - L| \geq \epsilon.$

It is sufficient to demonstrate that regardless of the proposed value of $L$, setting $\epsilon = 1$ will work.

With $L$ some fixed value, and $\epsilon = 1$, regardless of the positive value of $\delta$ chosen, you can arbitrarily decide that you are going to look for a (positive) $\delta_1 < \delta$
and then choose $x = 4 + \delta_1.$

Then, you will have that the numerator of $f(x)$ will be
greater than $8$, and
the denominator of $f(x)$ will be $\delta_1$.

Therefore, $f(x)$ will be greater than $\frac{8}{\delta_1}.$
Since you can choose $\delta_1$ as small as you want, you can choose $\delta_1$ small enough
so that $[f(x) - L] > 1 = \epsilon.$

This means that regardless of your proposed value of $L$, and regardless of your choice of (positive) $\delta$, you will be able to find at least one $x$ such that $0 < |x - 4| < \delta$ and
$|f(x) - L| > 1.$