Epsilon delta proof with fractions, given what delta must be Choose $\delta = \min (1, \frac{\epsilon}{10})$ 
is the following statement true?
$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$
Okay so this is what I have so far
$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}{|x^2+1|}$$
I then chose $\delta = \frac{1}{2}$ (because if I chose $\delta = 1$, $x$ would come out as $2$ which would give me $|x-1|\cdot 0 < \epsilon$ which doesn't tell me much?)
So if $\delta = \frac{1}{2}$, then
$\frac{1}{2}< x < \frac{3}{2}$ (because it's a fraction I used $x > \frac{1}{2}$)
$$|x-1|\cdot\frac{\left|-\frac{1}{2}+2\right|}{\left|(\frac{1}{2})^2+2\right|}=\frac{2}{3}|x-1|< \epsilon$$
$$|x-1|< \frac{3}{2}\epsilon$$
I'm just so confused because I don't know how to relate the value for $\delta$ (which I found to be $\frac{3}{2}\epsilon$ to the delta they're making me choose of $\min (1, \frac{\epsilon}{9})$. I'm also not sure if I chose the correct value for $x$ as it is a fraction. 

An alternative answer keeping the $\delta$:
We know that $|x-1|<\delta$ and 
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}$$
Now, we use the bounds
$$|x-2| = |x-1-1|\leq |x-1|+1 \leq\delta +1$$
and
$$|x^2+1|=x^2+1\geq 1$$
to obtain
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}\leq \delta(1+\delta) \leq \delta2 \leq \frac{\varepsilon}{5}<\varepsilon$$
using $\delta\leq \min\{1,\frac{\varepsilon}{10}\}$.
