Trouble with delta epsilon limit proof of form $\frac{a}{bx+c}$

Question

I want to prove $\lim_{x \to 2} \frac{3}{x+1} = 1 $ using the $ \delta $-$\epsilon$ definition. How can I select a $\delta $ based on $ \epsilon $ so this function stays bounded within 1?

Answer 1

Since $ lim (x) = 2 => |x-2| < 1 $ $\\$ => in this restricted domain, $ x $ belongs to $ (1, 3) $

Let's choose $\delta = 2\epsilon$.

$$ 0 < | x - 2| < \delta = 2\epsilon $$ $ $ $ => \left| \frac{3}{x+1} - 1 \right| = \left| \frac{2-x}{x+1} \right| = \frac{|x - 2|}{|x+1|}< \frac{|x-2|}{2} $ (since max($\frac{1}{|x+1|}) = \frac12$ in the above domain)

Finally, as $| x - 2| <\delta = 2\epsilon$, $$\left| \frac{3}{x+1} - 1 \right| < \epsilon$$

This method of restricting the domain can be used for any limit proof of the form $\frac{a}{bx+c}$.

Answer 2

Since the question is how to select a $\delta$ and not which delta works, I'll try to answer that, so it's more like a tutorial and won't give you the particular answer explicitly so that you find it yourself.

The problem is to prove that at $x=2$, $$\forall\varepsilon>0\exists\delta>0\forall x':\left|x'-x\right|\to\left|f(x')-f(x)\right|\le\varepsilon$$ where $f(x')=3/(x'+1)$. (I'll consider $x$ "fixed" and $x'$ "floating" in my answer)

1. Let $\varepsilon>0$ be given; now we have to find some $\delta>0$ such that (plugging $x=2$ and $f(x)$): $$\forall x':\left|x'-2\right|\leq\delta\to\left|\frac{3}{x'+1}-\frac{3}{2+1}\right|\le\varepsilon,$$ or more simply, $\forall x':\left|x'-2\right|\leq\delta\to\left|\frac{3}{x'+1}-1\right|\le\varepsilon$. The natural thing to do when we don't see a direct solution is to expand the definitions. Our target inequality is $\left|\frac{3}{x'+1}-1\right|\le\varepsilon$, and by definition it means $-\varepsilon\le\frac{3}{x'+1}-1\le\varepsilon$.

2. Now stop and think (so far we only did blind simplifications and definition expansions). We want the target equality fulfilled for all $x'$ sufficiently close to $x=2$, i.e. whenever $|x'-2|\le\delta$. So let's try to answer the question, for which $x'$ is it fulfilled? For which $x'$ does $-\varepsilon\le\frac{3}{x'+1}-1\le\varepsilon$? Solving the inequality for $x'$ we get $$\frac{3}{1+\varepsilon}-1\le x'\le\frac{3}{1-\varepsilon}-1.$$

3. Now stop and think again. The question is, what $\delta$ makes $|x'-2|\le\delta$ imply the target equality? We want $x'$ clamped in the interval $\left[\frac{3}{1+\varepsilon}-1,\frac{3}{1-\varepsilon}-1\right]$. And the question is to find an interval $[2-\delta,2+\delta]$ (corresponding to the $|x'-2|\le\delta$ condition) which fits within the target interval. Now that's a purely arithmetic question that I believe you'll be able to solve. And interestingly, it will give you also the best delta of all!