# 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?

##### 3 Answers

#### Explanation:

#lim_(xto-3)((x-3)cancel((x+3)))/cancel((x+3))=-3-3=-6#

#"there is a hole at x=- 3"#

#"the simplified version is linear with no hole"#

graph{x-3 [-12.66, 12.65, -6.33, 6.33]}

Limit is

#### 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

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

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

Hence evaluating the full thing gives

#### Explanation:

#"to compute limits of the form"#

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

#"when the numerator/denominator have leading"#

#"coefficients of 1"#

#• " numerator/denominator have equal degree (m = n) "#

#"then limit is 1"#

#• " degree of numerator "<" degree of denominator "#

#"then limit is 0"#

#• " degree of numerator ">" degree of denominator"#

#"then limit is "oo#

#"the limit does not exist if the right / left limits"#

#"are not equal"#