Proving limits for fractions using epsilon-delta definition Using the $\epsilon - \delta $ definition of the limit, prove that: $$\lim_{x\to 0} \frac{(2x+1)(x-2)}{3x+1} = -2$$
I firstly notice that my delta can never be greater than $\frac{1}{3}$ because there is a discontinuity at $x=-\frac{1}{3}$.
I applied the standard steps as follows:
$\vert \frac{(2x+1)(x-2)}{3x+1}  +2 \vert = \vert\frac{2x+3}{3x+1}\vert \vert x\vert$
Right now I need to restrict $x$ to some number, but I am not sure which value should I choose in order to easily bound my fraction, any help on choosing the correct delta is appreciated!
 A: Let $|x| \lt 1/5$, then
1)$-1/5< x < 1/5$, or 
$-3/5 +1< 3x +1<3/5+1$;
2) $|2x+3| \le 2|x| +3\lt 17/5$.
Let $\epsilon>0.$
Choose  $\delta = \min(1/5, (2/17)\epsilon )$, then 
$|x|\lt \delta$  implies
$\dfrac{|2x+3|}{|3x+1|}|x| \lt\dfrac{17/5}{2/5}|x| =$
$(17/2)\delta \lt \epsilon$.
A: Let $x > -\dfrac13$ and $|x| < \delta$, then $2x+3, 3x+1 > 0$, where $\delta > 0$ is to be determined.
$$\frac{2x+3}{3x+1} \le \frac{2\delta+3}{\underbrace{1-3\delta}_{\mbox{take $\delta < \frac13$}}}$$
Take $\delta < \dfrac13$ so that the denominator is positive.  Observe that when $|x| < \delta$, the fraction is positive, so the absolute sign can be omitted.
$$0<\frac{1}{1-3\delta} < 2 \iff 1-3\delta > \frac12 \iff \delta < \frac16 \implies \delta < \frac13$$
When $|x| < \delta < \dfrac16$, $2x + 3 < 2\delta + 3 < \dfrac{10}{3}$, so $\dfrac{2x+3}{3x+1} < \dfrac{20}{3}$.  If you want to cancel this factor in the final inequality, multiply $\epsilon$ with its inverse while defining $\delta$, i.e. set $\delta = \min\{\dfrac{3}{20} \epsilon, \dfrac16\}$.  When $|x| < \delta$,
$$\left\vert \frac{(2x+1)(x-2)}{3x+1}  +2 \right\vert = \frac{2x+3}{3x+1} \: \vert x\vert = \frac{20}{3} \cdot \dfrac{3}{20} \epsilon = \epsilon.$$
