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
Aug 26, 2017

-6

Explanation:

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

"there is a hole at x=- 3"

rArr"excluded value "x=-3

"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 rarr -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_(xrarr-3) (x^2-9)/(x+3) = lim_(xrarr-3) ((d)/(dx)(x^2-9))/((d)/(dx)(x+3))

lim_(xrarr-3)(2x)/1 = -6

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

lim_(xrarr-3) (x^2-9)/(x+3) = lim_(xrarr-3)((x+3)(x-3))/(x+3) = lim_(xrarr-3) x - 3 = -6

Hence evaluating the full thing gives

lim_(xrarr-3) (x^2-9)/(x+3) - 6 = -6 + lim_(xrarr-3) (x^2-9)/(x+3) = -6 - 6 = -12

Aug 26, 2017

"see explanation"

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"