Question

How do you evaluate the limit `(2x+4)/(x^2-3x-10)` as x approaches `2`?

Answer 1

`-2/3`

Explanation:

Since this function is not indeterminate at x = 2, we can evaluate it by direct substitution.

`lim_(xto2)(2x+4)/(x^2-3x-10)=(2(2)+4)/(2^2-3(2)-10)=8/(-12)=-2/3`

Answer 2

`lim_(xrarr2)(2x+4)/(x^2-3x-10) = -2/3` and `lim_(xrarr-2)(2x+4)/(x^2-3x-10) = -2/7`

Explanation:

As `x` approaches `2`, the numerator goes to `8` and the denominator goes to `4-6-10 = -12`, so the limit is `8/-12 = -2/3`

As `x` approaches `-2`, both the numerator and denominator go to `0`. Since boith are polynomials and `02` is a zero, we can be sure that `x-(-2)` = x+2# is a foctor of both the numerator and the denominator.

We'll factor and reduce.

`lim_(xrarr-2)(2x+4)/(x^2-3x-10)=lim_(xrarr-2)(2(x+2))/(x+2)(x-5))`

`= lim_(xrarr-2)2/(x-5)`

`= 2/((-2)-5) = -2/7`