Limit question: $\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$ 
Show that $$\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$$ by the definition.

This is what I've done so far:
$$\left\vert\frac{x+1}{x+2}  - \frac{3}{4} \right| = \left| \frac{x-2}{4(x+2)} \right|$$
and I see that
$$|x-a| = |x-2|$$ which is in our numerator. What choice of $\delta$ can I pick (and what is the motivation) so that it satisfies the following limit definition?
$$\lim_{x \rightarrow a} f(x) = b \Leftrightarrow \forall \epsilon >0, \exists \delta : |x-a| < \delta \Rightarrow |f(x) - b| < \epsilon$$

Note that 
$$
|\frac{x-2}{4(x+2)}| = \frac{|x-2|}{4|x+2|} 
$$
for all $x \neq -2$.
We have to "bound away" the annoying term $|x+2|$ by a preliminary upper bound for $|x-2|$. If $|x-2| < 1$, then $-1 < x-2 < 1$, so $1 < x < 3$, and hence $3 < x+2 < 5$; hence $3 < |x+2| < 5$. Then 
$$
\frac{|x-2|}{4|x+2|} < \frac{|x-2|}{4\cdot 3} = \frac{|x-2|}{12}.
$$
Given any $\varepsilon > 0$, we have $\frac{|x-2|}{12} < \varepsilon$ if $|x-2| < 12\varepsilon$. So combining the two upper bounds for $|x-2|$ we conclude that $\delta := \min \{ 1, 12\varepsilon \}$ suffices.
