# Evaluate the Limits?

## Evaluate the limits A) ${\lim}_{x \to 2} \left(\frac{3 {x}^{2} - 7 x + 2}{x - 2}\right)$ B) ${\lim}_{x \to - 3} \frac{x + 3}{\sqrt{{x}^{2} - 5} - 2}$ C) ${\lim}_{x \to - 2} \frac{{x}^{3} + {x}^{2} - 8 x - 12}{{x}^{3} + 8 {x}^{2} + 20 x + 16}$

Apr 27, 2018

A) $5$

B) $- \frac{2}{3}$

C) $- \frac{5}{2}$

#### Explanation:

One way to do this is to graph this and look at the the function as it approaches the desired limits so in this case for the first one as $x$ is nearly at $2$, $x$ nearly at $- 3$ and $x$ at nearly $- 2$

If you want the more specific way then do this:

A) The first one is solved by factoring because if you directly substitute the number $2$ in for the $x$'s you'll see you get $\frac{0}{0}$ which you can't have.

Factor the numertor:

${\lim}_{x \to 2} = \left(\frac{\left(x - 2\right) \left(3 x - 1\right)}{x - 2}\right)$

${\lim}_{x \to 2} = \left(\frac{\left(\cancel{x - 2}\right) \left(3 x - 1\right)}{\cancel{\left(x - 2\right)}}\right)$

${\lim}_{x \to 2} = 3 x - 1$

now subsitute the $2$ into $x$ in the equation:

${\lim}_{x \to 2} = 5$

B) This one also gives $\frac{0}{0}$ if you plug-in the $- 3$ straight away.

Flip the numerator and denominator

${\lim}_{x \to - 3} = \frac{x + 3}{\sqrt{{x}^{2} - 5} - 2}$

${\lim}_{x \to - 3} = \frac{\sqrt{{x}^{2} - 5} + 2}{x - 3}$

Plug-in $x = - 3$

${\lim}_{x \to - 3} = - \frac{2}{3}$

C) If you plug-in right away you get the same thing $\frac{0}{0}$

Here what you do is take the derivative of both numerator and denominator seperately because you have an ${x}^{3}$ on both numerator and denominator

Derivative of numerator:

$\left({x}^{3} + {x}^{2} - 8 x - 12\right) \frac{d}{\mathrm{dx}}$

use the power rule and you get:

$3 {x}^{2} + 2 x - 8$

Derivative of denominator:

$\left({x}^{3} + 8 {x}^{2} + 20 x + 16\right) \frac{d}{\mathrm{dx}}$

using power rule again:

$3 {x}^{2} + 16 x + 20$

Now to evaluate the limit by plugging-in $x = - 2$:

${\lim}_{x \to - 2} = \frac{3 {x}^{2} + 2 x - 8}{3 {x}^{2} + 16 x + 20}$

${\lim}_{x \to - 2} = - \frac{5}{2}$