Epsilon-Delta proof for a limit of a function

Question

$\lim\limits_{x \to 2} \frac{x^2+4}{x+2}=2$

I understand the structure of the epsilon delta proof, but I need help with the scratchwork/setup.

Using |$\frac{x^2+4}{x+2}-2$| $<\epsilon$ , you can factor and you're left with |$x-4$|$< \epsilon$. What I'm stuck on is solving for $x-a$ or $x-2$.

Can I do something with the Triangle Inequality to say |$x-2-2$| $\le$ |$x-2$|$ \;+\; 2$ $< \epsilon$

Any help would be appreciated!

Answer 1

Since\begin{align}\left|\frac{x^2+4}{x+2}-2\right|&=\left|\frac{x^2-2x}{x+2}\right|\\&=\frac{|x|}{|x+2|}|x-2|,\end{align}you can observe first that $|x-2|<1\implies|x|\leqslant|x-2|+2<3$ and $|x+2|=\bigl|4-(-x+2)\bigr|>3$. Therefore$$\frac{|x|}{|x+2|}<1.$$So, if you fix $\varepsilon>0$ and if you take $\delta=\min\{1,\varepsilon\}$, then$$\frac{|x|}{|x+2|}|x-2|<\varepsilon.$$

Answer 2

$$\left|\frac{x^2+4}{x+2} -2 \right|=\left|\frac{x^2-2x}{x+2} \right|=\frac{|x||x-2|}{|x+2|}$$

WLOG, we can assume that $\delta < 1$,

Hence if $|x-2| < \delta$, then $2-\delta < x < 2+\delta$ which implies that $x$ is between $1$ and $3$ while $x+2$ is in between $3$ and $5$.

Hence, $$\left|\frac{x^2+4}{x+2} -2 \right|\leq \frac{3|x-2|}{3}=|x-2|$$

Remark about your triangle inequality approach:

what if $\epsilon < 2$.