<s>[INST] Epsilon-Delta proof for a limit of a function $\lim\limits_{x \to 2} \frac{x^2+4}{x+2}=2$
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!
 [/INST] $$\left|\frac{x^2+4}{x+2} -2 \right|=\left|\frac{x^2-2x}{x+2} \right|=\frac{|x||x-2|}{|x+2|}$$
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 $\epsilon < 2$.
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