3 years ago

Kner Page Classes

Mathematics

1 answer:

Ann [662]3 years ago

7 0

Answer:

Step-by-step explanation:

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Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min

leva [86]

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

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

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