Proving limit using $\epsilon$ and $\delta$

Question

How can I prove using the formal definition of a limit (epsilon delta) that the limit of $(x+1)/(x+2)$ as $x$ approaches $2$ is $3/4$? I am aware that you can use continuity arguments to prove this limit also, but I am trying to do this using the definition.

I tried to reverse engineer the thing which got me to $|x-2|< 4\epsilon|(x-2)+4|$. I then tried to square both sides hoping to get some kind of bound on $(x-2)$, and while I did, it turned out to be quite a complicated expression which meant that too much time would be wasted in an exam. Also, it would be difficult to justify the choice of delta if I did this.

Is there a shorter way of approaching this problem?

Answer 1

Take $\delta\leq 1$ (we may replace $1$ with any positive number less than 4 in order to stay away from the pole $x=-2$). Then $|x-2|<\delta$ implies that $x\in (1,3)$ and $3 < |x+2|< 5$. Hence, for $\epsilon>0$, $$\left|\frac{x+1}{x+2}-\frac{3}{4}\right|=\frac{|x-2|}{4|x+2|}\leq \frac{|x-2|}{4\cdot 3}<\delta\leq \epsilon$$ with $\delta=\min(1,\epsilon)$.

Answer 2

Pick $\delta=\min\{1,\epsilon\}$. Here is why: $$\left|\frac{x+1}{x+2}-\frac34\right|=\frac{\left|{x-2}\right|}4\frac{1}{\left|{x+2}\right|}$$ We can restrict $\delta\leq1$ so that $|x-2|<1\implies 1<x<3$ so $\,3<|x+2|<5\implies \frac13>\frac1{|x+2|}>\frac15$, thus $$\frac{\delta}4\frac{1}{\left|{x+2}\right|}\leq\frac{\delta}4\cdot\frac{1}{3}=\frac{\delta}{12}<\epsilon.$$ Since $\epsilon$ is a small positive number, we have that $\epsilon<12\epsilon. $

In this case, it is not important if you bounded $\delta$ with $1,2$ or any other number. Just make sure that you adjust everything else.