Question

Question #84e30

Answer 1

As presented, the limit does not exist.

Explanation:

As `xrarr-3`, the numerator goes to `sqrt10` and the denominator goes to `0`, so the absolute value of the ratio is increasing without bound.

The limit as `xrarr-3^+` has the numerator approaching `sqrt10` while the denominator goes to `0` through positive values. The ratio increases without bound.

That is `lim_(xrarr-3^+) (sqrt(x^2-x-2)-2)/(x+3) = oo`

From the other side, as `xrarr-3^-` has the numerator approaching `sqrt10` while the denominator goes to `0` through negative values. The ratio decreases without bound.

That is `lim_(xrarr-3^-) (sqrt(x^2-x-2)-2)/(x+3) = -oo`.

The two-sided limit simply does not exist.

Answer 2

If the intended problem is `lim_(xrarr-3^+)(sqrt(x^2+x-2)-2)/(x+3)`, then the limit is `-5/4`

Explanation:

The initial form of `lim_(xrarr-3^+)(sqrt(x^2+x-2)-2)/(x+3)` is the indeterminate `0/0`. Try rationalizing the numerator.
(You may not be sure it will work, but you need to try something. I am sure if will work because I've done lots of limits like this.)

`lim_(xrarr-3*+)(sqrt(x^2+x-2)-2)/(x+3) = lim_(xrarr-3^+)((sqrt(x^2+x-2)-2))/((x+3)) ((sqrt(x^2+x-2)+2))/((sqrt(x^2+x-2)+2))`

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

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

Note that we still get `0/0`, but we can factor and reduce now.

`= lim_(xrarr-3^+)((x+3)(x-2))/((x+3) (sqrt(x^2+x-2)+2))`

`= lim_(xrarr-3^+)(x-2)/ (sqrt(x^2+x-2)+2)`

`= ((-3)-2)/(sqrt((-3)^2+(-3)-2)+2)`

`= (-5)/(sqrt4+2) = (-5)/4`