"Find the limits: limx→−2x+2x2+5−3" was answered by students like you

UkusakazaL

Answered question

2021-03-09

Find the limits:

lim_{x \to - 2} \frac{x + 2}{\sqrt{x^{2} + 5} - 3}

Answer & Explanation

nitruraviX

Skilled2021-03-10Added 101 answers

Given limits:
$lim_{x \to - 2} \frac{x + 2}{\sqrt{x^{2} + 5} - 3}$
Rationalize the denominator, we get
$lim_{x \to - 2} \frac{x + 2}{\sqrt{x^{2} + 5} - 3} \times \frac{\sqrt{x^{2} + 5} + 3}{\sqrt{x^{2} + 5} + 3}$
$lim_{x \to - 2} \frac{(x + 2) (\sqrt{x^{2} + 5} + 3)}{(\sqrt{x^{2} + 5})^{2} - 3^{2}}$
$lim_{x \to - 2} \frac{(x + 2) (\sqrt{x^{2} + 5} + 3)}{x^{2} + 5 - 9}$
$lim_{x \to - 2} \frac{(x + 2) (\sqrt{x^{2} + 5} + 3)}{x^{2} - 4}$
$lim_{x \to - 2} \frac{(x + 2) (\sqrt{x^{2} + 5} + 3)}{x^{2} - 2^{2}}$
$lim_{x \to - 2} \frac{(x + 2) (\sqrt{x^{2} + 5} + 3)}{(x - 2) (x + 2)}$
$lim_{x \to - 2} \frac{(\sqrt{x^{2} + 5} + 3)}{(x - 2)}$
Apply the limits $x = - 2$ , we get
$lim_{x \to - 2} \frac{(\sqrt{x^{2} + 5} + 3)}{(x - 2)} = \frac{(\sqrt{(- 2)^{2} + 5} + 3}{(- 2 - 2)}$
$lim_{x \to - 2} \frac{(\sqrt{x^{2} + 5} + 3)}{(x - 2)} = \frac{(\sqrt{4 + 5} + 3}{(- 2 - 2)}$
$lim_{x \to - 2} \frac{(\sqrt{x^{2} + 5} + 3)}{(x - 2)} = \frac{(3 + 3)}{(- 2 - 2)}$
$lim_{x \to - 2} \frac{(\sqrt{x^{2} + 5} + 3)}{(x - 2)} = - \frac{3}{2}$

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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