# How do you determine the limit of (x+2)/(3x^2) as x approaches 0-?

May 18, 2018

${\lim}_{x \to 0 -} \frac{x + 2}{3 {x}^{2}} = \infty$

#### Explanation:

I would do it this way:
Let $x = \frac{1}{y}$ with $x < 0$,

$\frac{x + 2}{3 {x}^{2}}$ = $\frac{\frac{1}{y} + 2}{\frac{3}{y} ^ 2}$

=$\frac{y + 2 {y}^{2}}{3}$ (I multiplied numerator and denominator with ${y}^{2}$ to get rid of the fractions in the numerator and denominator)

We see that ${\lim}_{y \to - \infty} \frac{y + 2 {y}^{2}}{3} = \infty$ since ${\lim}_{y \to - \infty} \frac{2}{3} {y}^{2} = \infty$

Therefore ${\lim}_{x \to 0 -} \frac{x + 2}{3 {x}^{2}} = \infty$

This makes sense, since as $x \to 0 -$ the numerator will be closer and closer to 2, i.e.

$\frac{x + 2}{3 {x}^{2}} \to \frac{2}{3 {x}^{2}} \to \frac{2}{0}$ which has $\infty$ as it's limit.