Proving limits for fractions using epsilon-delta definition

Question

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!

Answer 1

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

$\dfrac{|2x+3|}{|3x+1|}|x| \lt\dfrac{17/5}{2/5}|x| =$

$(17/2)\delta \lt \epsilon$.

Answer 2

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 $|x| < \delta$,

$$\left\vert \frac{(2x+1)(x-2)}{3x+1} +2 \right\vert = \frac{2x+3}{3x+1} \: \vert x\vert = \frac{20}{3} \cdot \dfrac{3}{20} \epsilon = \epsilon.$$