Question

Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x minus 1 )squared minus 75 Over x plus 2 EndFraction

Answer 1

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

                                $\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2}$.

First, we simplify the expression in the numerator of the fraction.

$3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72$

Now, we obtain

                         $12(x^2-x-6) = 12(x+2)(x-3)$

and the fraction is transformed into

                       $\frac{3(2x-1)^2 - 75}{x+2} = \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)$

Therefore, the following limit is

       $\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2} 12(x-3) = 12 \lim_{x \to -2} (x-3)$

You can plug in $-2$ in the equation, hence

                        $12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60$