# How do you find lim (y+1)/((y-2)(y-3)) as x->3^+?

Dec 6, 2017

${\lim}_{y \to {3}^{+}} f \left(y\right) = \infty$

#### Explanation:

let $f \left(y\right) = \frac{y + 1}{\left(y - 2\right) \left(y - 3\right)}$

we know as $y \to {3}^{+} , y + 1 \to 4$
we know as $y \to {3}^{+} , y - 2 \to 1$
we know as $y \to {3}^{+} , y - 3 \to 0$

for this problem we are going to write $f \left(y\right)$ as;
$\frac{y + 1}{y - 2} \frac{1}{y - 3} = f \left(y\right)$

we know that $\frac{y + 1}{y - 2} \to 4$ as $y \to {3}^{+}$

But now we must consider$\frac{1}{y - 3}$

as $y \to {3}^{+}$ the demominator gets very small $\left(\to 0\right)$ but as $y \to {3}^{+}$ it is possotive, so hence as denominator $\to 0$ , $\frac{1}{y - 3} \to \infty$

So hecne $4 \cdot \infty = \infty$

= $\infty$