Epsilon delta proof with fractions, given what delta must be

Question

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.

Answer 1

As $|x-1|<1$ for all possible $\delta$ values then $$|x-2|< 2\ \text{and} \ |x^2+1|>1$$ So then we have $$|x-1|\frac{|-x+2|}{|x^2+1|}< \frac{2|x-1|}{1} = 2|x-1|$$

If $\epsilon\geq10$ then remembering $|x-1|<1$ as $\delta =1$ in this case $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot 1<10\leq\epsilon$$ as desired. If $\epsilon<10$ then remembering $|x-1|<\delta=\frac{\epsilon}{10}$ we have $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot \frac{\epsilon}{10}=\epsilon/5<\epsilon$$ and you are finished. Let me know if you have questions.

Answer 2

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 $\delta\leq \min\{1,\frac{\varepsilon}{10}\}$.