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"