# epsilon delta approach to a problem

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 $$2\epsilon$$

• You're working backwards. You want to exhibit a $\delta$ in terms of $\epsilon$ that ensures that $\frac{|x-5|}{|x-3|}<\epsilon$ whenever $|x-5|<\delta$. Anyway, to answer your question, $|x-3|=x-3$ here, because $x$ is only considered close to $5$ (see the limit). Also note that you're missing a $\frac{1}{2}$. It should be $\frac{|x-5|}{2|x-3|}$.
– Doc
Commented Nov 26, 2013 at 0:31
• Why don't you try to use the triangle inequality on $\vert x-3 \vert = \vert (x-5)+2 \vert$. That might help a bit with the denominator. Commented Nov 26, 2013 at 0:36
• I understood the part that you suggested. So, would it be wrong ? or is this an acceptable answer ? Commented Nov 26, 2013 at 1:32

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 $$\left|\frac1{x-3}-\frac12\right|<\frac\epsilon{|x-3|}\le\epsilon.$$
• You are correct that if $|5-x|<1,$ then $|x-3|<3.$ However, this does not help you, as it implies that $\frac1{|x-3|}>\frac13,$ and this inequality is going the wrong direction. More important is the fact that if $|5-x|<1,$ then $|x-3|>1,$ so $\frac1{|x-3|}<1.$ This is why we chose $\delta=\min\{1,2\epsilon\}$ rather than $\delta=\min\{1,6\epsilon\}.$ Commented Nov 26, 2013 at 12:12
• As a concrete example to see that your approach fails, take $\epsilon=\frac16,$ so that $\delta=1$ (in your approach). Now, let $x=\frac{17}4,$ so that $|x-5|=\frac34<1=\delta,$ but $$\frac{|x-5|}{2|x-3|}=\cfrac{\frac34}{\frac52}=\frac3{10}=\frac{18}{60}> \frac{10}{60}=\frac16=\epsilon.$$ Commented Nov 26, 2013 at 15:07