epsilon delta approach to a problem

Question

The following is what I would like to show.

$$\lim_{x \to5}\frac{1}{x-3}=1/2$$

Given any $\epsilon>0$, $|\frac{1}{x-3}-\frac{1}{2}|\le\frac{|x-5|}{|x-3|}\le2\epsilon$

How do I get rid of $|x-3|$ here to get to

$|x-5|<\delta$ ?

My friends told me something about $\delta=\min\{{1,2\epsilon\}}$ being a part of the answer, but I did not get it.

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This is how I would do it.

We are given that $$|\frac{1}{x-3}-\frac{1}{2}|<\epsilon$$ we want to have

$$|5-x|<2\epsilon|x-3|\le \delta$$.

Since $|5-x|<1$ implies $|x-3|<3$, we can choose $\delta=\min\{1,6\epsilon\}$

because if $\delta=1$ then $1<6\epsilon$ so

$$|\frac{1}{x-3}-\frac{1}{2}|=\frac{|x-5|}{2|x-3|}<\frac{1}{6}<\epsilon$$

and if $\delta=6\epsilon$ then

$$|\frac{1}{x-3}-\frac{1}{2}|=\frac{|x-5|}{2|x-3|}<\frac{6\epsilon}{6}<\epsilon$$.

Therefore $$\lim_{x \to5}\frac{1}{x-3}=1/2$$.

Does this look okay ?

I am still a bit worried because I got $6\epsilon$ instead of $2\epsilon$

Answer 1

We start by noting that $$\left|\frac1{x-3}-\frac12\right|=\left|\frac{2}{2(x-3)}-\frac{x-3}{2(x-3)}\right|=\left|\frac{5-x}{2(x-3)}\right|=\frac12\cdot\frac{|x-5|}{|x-3|}$$

We're going to be looking for an upper bound for $|x-5|$ already, and the constant won't give us any trouble, but we need to do something about that $|x-3|.$ In order to ensure an upper bound on the far-right expression above, we need to ensure a lower bound for $|x-3|.$ Fortunately, we're letting $x\to 5,$ so that won't be a problem. In particular, when $|x-5|\le 1,$ we have $x\ge 4,$ so that $|x-3|=x-3\ge 1,$ and so $\frac1{x-3}\le 1.$ Hence, if we put $\delta=\min\{1,2\epsilon\},$ then we have for all $x$ such that $|x-5|<\delta$ that $$\left|\frac1{x-3}-\frac12\right|=\frac12\cdot\frac{|x-5|}{|x-3|}<\frac\epsilon{|x-3|}$$ since $|x-5|<\delta\le2\epsilon,$ and so since $|x-5|<\delta\le1,$ then $$\left|\frac1{x-3}-\frac12\right|<\frac\epsilon{|x-3|}\le\epsilon.$$