Question

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?

Answer 1

`-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]}

Answer 2

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`

Answer 3

`"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"`