# Why is lim_(xrarr-3)(x^2-9)/(x+3) -6? If I solve it by factoring, yes I get -6 but I read somewhere that if the degree of the numerator is greater than the degree of the denominator, then the limit does not exist?

Aug 26, 2017

$- 6$

#### Explanation:

${\lim}_{x \to - 3} \frac{\left(x - 3\right) \cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)}} = - 3 - 3 = - 6$

$\text{there is a hole at x=- 3}$

$\Rightarrow \text{excluded value } x = - 3$

$\text{the simplified version is linear with no hole}$
graph{x-3 [-12.66, 12.65, -6.33, 6.33]}

Aug 26, 2017

Limit is $- 12$.

#### Explanation:

Ok the "-6" term is constant and so doesn't change as we vary x so we shall ignore it for now.

The problem isn't to do with the degrees of the numerator/denominator, the problem is that as $x \rightarrow - 3$ both numerator and denominator go to zero. Hence the limit is in indeterminate form. Can do this in a few ways"

In general I tend to go with L'Hopital's rule.

${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - 9}{x + 3} = {\lim}_{x \rightarrow - 3} \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} - 9\right)}{\frac{d}{\mathrm{dx}} \left(x + 3\right)}$

${\lim}_{x \rightarrow - 3} \frac{2 x}{1} = - 6$

However in this case it is simpler to just factorise the numerator:

${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - 9}{x + 3} = {\lim}_{x \rightarrow - 3} \frac{\left(x + 3\right) \left(x - 3\right)}{x + 3} = {\lim}_{x \rightarrow - 3} x - 3 = - 6$

Hence evaluating the full thing gives

${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - 9}{x + 3} - 6 = - 6 + {\lim}_{x \rightarrow - 3} \frac{{x}^{2} - 9}{x + 3} = - 6 - 6 = - 12$

Aug 26, 2017

$\text{see explanation}$

#### Explanation:

$\text{to compute limits of the form}$

•color(white)(x)lim_(xtooo)(x^n+" lower power terms")/(x^m+" lower power terms")

$\text{when the numerator/denominator have leading}$
$\text{coefficients of 1}$

• " numerator/denominator have equal degree (m = n) "
$\text{then limit is 1}$

• " degree of numerator "<" degree of denominator "
$\text{then limit is 0}$

• " degree of numerator ">" degree of denominator"
$\text{then limit is } \infty$

$\text{the limit does not exist if the right / left limits}$
$\text{are not equal}$