# How do you find the Limit of lim_(yto3)(y^3-13y+12 )/( y^3-14y+15)?

Jul 10, 2017

Because the expression evaluated at the limit results in the indeterminate form $\frac{0}{0}$, the use of L'Hôpital's rule is warranted.

#### Explanation:

Given: ${\lim}_{y \to 3} \frac{{y}^{3} - 13 y + 12}{{y}^{3} - 14 y + 15}$

Use L'Hôpital's rule:

${\lim}_{y \to 3} \frac{\frac{d \left({y}^{3} - 13 y + 12\right)}{\mathrm{dx}}}{\frac{d \left({y}^{3} - 14 y + 15\right)}{\mathrm{dx}}} =$

${\lim}_{y \to 3} \frac{3 {y}^{2} - 13}{3 {y}^{2} - 14} = \frac{14}{13}$

So goes the original limit:

${\lim}_{y \to 3} \frac{{y}^{3} - 13 y + 12}{{y}^{3} - 14 y + 15} = \frac{14}{13}$

Jul 10, 2017

$\frac{14}{13.}$

#### Explanation:

The Nr. =${y}^{3} - 13 y + 12.$ and, the Dr. =${y}^{3} - 14 y + 15.$

Note that, in the polynomial of the Nr., the sum of the co-effs. is $0.$

$\therefore \left(y - 1\right)$ must be a Factor.

$\therefore \text{ The Nr.} = \left(y - 1\right) \left({y}^{2} + y - 12\right) = \left(y - 1\right) \left(y + 4\right) \left(y - 3\right) .$

The poly. in the Dr. vanishes for $y = 3. \therefore \left(y - 3\right)$ is a Factor.

$\text{ The Dr. } = \left(y - 3\right) \left({y}^{2} + 3 y - 5\right) .$

$\therefore \text{ The Lim. } = {\lim}_{y \to 3} \frac{\cancel{\left(y - 3\right)} \left(y - 1\right) \left(y + 4\right)}{\cancel{\left(y - 3\right)} \left({y}^{2} + 3 y - 5\right)} ,$

$= {\lim}_{y \to 3} \frac{\left(y - 1\right) \left(y + 4\right)}{{y}^{2} + 3 y - 5} ,$

$= \frac{\left(3 - 1\right) \left(3 + 4\right)}{{3}^{2} + 3 \cdot 3 - 5} ,$

$\Rightarrow \text{ The Lim. } = \frac{14}{13.}$