# Question 55f2d

Sep 9, 2016

${\lim}_{x \rightarrow 2} \frac{{x}^{3} - 3 {x}^{2} + 4}{{x}^{3} - {x}^{2} - 8 x + 12} = \textcolor{g r e e n}{\frac{3}{5}}$

#### Explanation:

The problem is that both:
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} - 3 {x}^{2} + 4$ and
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} - {x}^{2} - 8 x + 12$
are equal to $0$ if $x = 2$

...but this means that both of these expressions are divisible by $\left(x - 2\right)$

and using either synthetic or long division we can get:
$\textcolor{w h i t e}{\text{XXX")(x^3-3x^2+4)/(x^3-x^2-8x+12)=(cancel(""(x-2))(x^2-x-2))/(cancel(} \left(x - 2\right)} \left({x}^{2} + x - 6\right)$

Unfortunately both
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - x - 2$ and
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x - 6$
are also both equal to $0$ if $x = 2$

...but (again) this means that both of these expressions are divisible by $\left(x - 2\right)$

and we can get
color(white)("XXX")(x^2-x-2)/(x^2+x-6)=(cancel(""(x-2))(x+1))/(cancel(""(x-2))(x+3))#

So we have
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow 2} \frac{{x}^{3} - 3 {x}^{2} + 4}{{x}^{3} - {x}^{2} - 8 x + 12} = {\lim}_{x \rightarrow 2} \frac{x + 1}{x + 3}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXX}} = \frac{2 + 1}{2 + 3} = \frac{3}{5}$

Sep 9, 2016

I found: $\frac{3}{5}$

Try this: