Need help proving continuity for $\frac{2x}{3x-1}$ at $x=1$ using Epsilon-Delta

Question

The function is given as:

$\frac{2x}{3x-1}$

I must prove continuity at $x=1$.

I understand the definitions using the Epsilon Delta proof but my issue comes when I arrive at

$|f(x)-f(1)|=|\frac{1-x}{3x-1}|=\frac{|x-1|}{|3x-1|}$< $ \epsilon$.

I know I must relate that result with $|x-1| < \delta $ but I am stumped on how to bound the denominator of the first result $|3x-1|$

I think I may have figured out my issue, I have been setting $\delta < 1$ but that is not sufficient and I have to go further, I instead and say what if $\delta < 1/3$

In that case I end up with $|3x-1|>1$ so I can go back to my original result to get, $\frac{|x-1|}{|3x-1|} < \frac{|x-1|}{1} = |x-1| < \delta = \epsilon$. So in other words I suppose I should set $\delta = \epsilon$? Or $\delta = min(\epsilon, 1/3)$

Answer 1

A common theme of explicit epsilon-delta calculations is having a "part" of the formula for the difference that isn't helping the difference to get small in the limit in question. In this case that is the division by $|3x-1|$; since $|3x-1|$ isn't blowing up, this division isn't helping you. So you generally want to replace that part by something simpler by imposing a restriction on the values of $\delta$ that doesn't depend on the chosen value of $\epsilon$.

Here you can say that if $x$ stays away from $1/3$ then $\frac{1}{|3x-1|}$ can be bounded by a constant. Now the only way to keep $x$ away from some point is to keep it close to $1$. So this forces $\delta<2/3$, so that $|3x-1|$ is bounded away from zero by some constant. To get a nice explicit bound you can pick whatever cap for $\delta$ that is strictly less than $2/3$ that you want; here the algebra is convenient if you pick $\delta \leq 1/3$ so that $x>2/3$ and so $|3x-1|>1$. But there was not really anything special about $1/3$, we could have used any number between $0$ and $2/3$ for our absolute cap for $\delta$.

From this discussion we conclude that $|f(x)-f(1)|<|x-1|$ if $|x-1|<1/3$, and so now you can indeed take $\delta=\min \{ 1/3,\epsilon \}$ to complete the proof of continuity.

Answer 2

Observe that if $\;x\;$ is very close ot $\;1\;$, say $\;0.9\le x\le1.1\;$ , then

$$2.7\le3x\le3.3\implies1.7\le3x-1\le2.3\implies 3x-1>\frac32\,,\,\,\,\text{say}$$

and we thus get:

$$|f(x)-f(1)|=\frac{|x-1|}{|3x-1|}\le\frac32|x-1|<\frac32\delta\stackrel{\text{we want!}}<\epsilon\implies\delta<\frac{2\epsilon}3$$

Thus, it is enough to take $\;\delta=\min\left\{0.1,\,\frac{2\epsilon}3\right\}\;$, for any $\;\epsilon>0\;$ , to get what we want.