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Determine whether $(4x-3)/(x-2)$ is uniformly continuous on the open interval $(1,2)$. I'm not sure how to start this as I have only answered these questions with closed intervals?

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Well it's worth looking back at the definition of uniform continuity. It says $f(x)$ is continuous on an interval $I$ if $$\forall(\epsilon>0)\exists (\delta>0)\forall (x,y\in S)\;\;(|x-y|<\delta)\implies (|f(x)-f(y)|<\epsilon$$

This means that for every $\epsilon$ you can find a single $\delta$ such that for any two points on the interval if their distance is less than $\delta$ then the distance of their images is closer then $\epsilon$. This sound a lot like saying that the rate of change of the function $f$ can be bounded (I don't want to say it's the same because I'm not sure, but it is a decent way to think about the condition in my opinion. When I have more time I will think about it and correct if necessary with counter examples).

This intuition should give you an idea of whether or not this function can be uniformly continuous on the interval proving one way or another is not that hard once you decide what to prove I think.

Hint: Look at how the function behaves close to $2$.

Ok full proof follows.

We will show that (4x-3)/(x-2) is not uniformly continuous on the open interval (1,2). The obvious problem is at the point $2$ where the function has a vertical assymptote.

Let $\epsilon=1$, let $\delta$ be arbitrary fixed. Notice that close to $2$ the function behaves roughly like $5/(x-2)$. Assume that $2>y>x>3/4$, then $(4y-3)>(4x-3)>0$, $0>(y-2)>(x-2)$ and $(4x-3)/(x-2)-(4y-3)/(y-2)>0$. Next $$\frac{4x-3}{x-2}-\frac{4y-3}{y-2}=\frac{(4x-3)(y-2)-(4y-3)(x-2)}{(x-2)(y-2)}$$

We can see what's going on here. Since we will want that last fraction to be large ( greater then 1), we will want to bound the top from below. Let's look at the numerator only now:

$$(4x-3)(y-2)-(4y-3)(x-2)=4xy-3y-8x+6-(4xy-8y-3x+6)=-8(x-y)+3(x-y)=-5(x-y)$$

So we now have:

$$\frac{(4x-3)(y-2)-(4y-3)(x-2)}{(x-2)(y-2)}=\frac{-5(x-y)}{(x-2)(y-2)}$$

Now we know that we need $|x-y|<\delta$, so we might as well choose $y-x=\min\{\delta/2,1/4\}$, that will ensure that we satisfy the left side of our implication. We will call the number chosen $\sigma$. We then have $$\frac{-5(x-y)}{(x-2)(y-2)}=\frac{5\sigma}{(x-2)(x+\sigma-2)}$$

Remember we have $$\frac{4x-3}{x-2}-\frac{4y-3}{y-2}=\frac{5\sigma}{(x-2)(x+\frac{\delta}{2}-2)}$$ Next we plce the absolute values in. We use the usual rules about products of absolute values being absolute values of products etc. That gives us: $$\left|\frac{4x-3}{x-2}-\frac{4y-3}{y-2}\right|=\frac{5\sigma}{|x-2||x+\sigma-2|}$$

Now we want to bound the expression from below so things in the numerator must get bounded from above. Let us then choose $x$ so that $|x-2|<\sigma$. Remember that $\sigma\leq 1/4$ then we get:

$$\frac{5\sigma}{|x-2||x+\sigma-2|}\geq \frac{5\sigma}{\sigma(\sigma+\sigma)}=\frac{5}{2\sigma}\geq\frac{5}{1/2}=10>1=\epsilon$$

The next to last of these inequalites comes from having chose $\sigma\leq 1/4$.So notice that given an $\epsilon=1$ for any $\delta$ we can find $x,y$ no further apart then $\delta$ such that the difference of the function values is bigger then $\epsilon$. Q.E.D.

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  • $\begingroup$ I realise that there would be an asymptote at 2 and the graph would approach this asymptote from the left in the given interval but I don't see how this helps? And we haven't covered anything to do with the derivative yet so I can't do it that way. $\endgroup$ – Alex Mar 6 '15 at 12:21
  • $\begingroup$ @alex You don't need the derivative for this though it might help (surprisingly it doesn't help as much as I thought). I would recommend math.wisc.edu/~robbin/521dir/cont.pdf as a great explanation btw. But to give a further hint, You say you know the function has an asymptote at $2$. This means it becomes very big very fast (it's a vertical asymptote). Consider how the values of $f(x),f(y)$ close together but also close to $2$ will behave? Can you see why you might have a problem bounding $|f(x)-f(y)|$? If so consider trying to show they can't be bounded. $\endgroup$ – DRF Mar 6 '15 at 12:43
  • $\begingroup$ For counterexamples see e.g. this answer and answers to this question. $\endgroup$ – Ruslan Mar 6 '15 at 13:13
  • $\begingroup$ Looking at my definitions, to prove it isn't uniformly continuous (as I don't think it is) I have to find 2 sequences Xn and Yn in the interval such that the modulus of their difference tends to 0 as n tends to infinity...but this is the bit I can't seem to do! $\endgroup$ – Alex Mar 6 '15 at 13:23
  • $\begingroup$ @Alex Hmm that seems to complicated. You can just, choose $\epsilon$ smartly, take $\delta$ arbitrary and show that you can find a pair $x,y$ such that $|x-y|<\delta$ but $|f(x)-f(y)|>\epsilon$. I can write up a more detailed proof in the answers later. $\endgroup$ – DRF Mar 6 '15 at 14:00
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We can find sequences $(x_n)$ and $(y_n)$ in $(1,2)$ such that $\lim_{n \to \infty}|x_n-y_n| = 0$ and $\lim_{n \to \infty}|f(x_n)-f(y_n)| \neq 0.$

For $n > 1$, take $x_n = 2- 1/2n$ and $y_n = 2-1/n$. Then $|x_n-y_n| = 1/2n \to 0$.

We have

$$\begin{align}|f(x_n) -f(y_n)| &= \left|\frac{4x_n-3}{x_n-2}-\frac{4y_n-3}{y_n-2}\right| \\ &=\frac{|(y_n-2)(4x_n-3)-(x_n-2)(4y_n-3)|}{|x_n-2||y_n-2|}\\ &= 2n^2\left|(y_n-2)(4x_n-3)-(x_n-2)(4y_n-3)\right|\\&=2n^2\left|-(1/n)(5-2/n)+(1/2n)(5-4/n)\right|\\ &= 2n^2\left|- 5/(2n)\right|\\ &= 5n\end{align}.$$

Hence, $|f(x_n)-f(y_n)| \to \infty$ and the function is not uniformly continuous.

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  • $\begingroup$ You're welcome. It appears you were looking for the sequential approach. $\endgroup$ – RRL Mar 10 '15 at 14:39

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