Question

problem. 19 : calculate the following limit. (if the limit does not exist, enter dne.)
\lim_{x \to -1} \frac{\sqrt{x^2 + 8} - 3}{x + 1} =

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x^2 + 8} + 3$.
$$\lim_{x \to -1} \frac{\sqrt{x^2 + 8} - 3}{x + 1} \cdot \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3}$$
Using the difference of squares formula $(a - b)(a + b)=a^2 - b^2$ in the numerator, we get:
$$\lim_{x \to -1} \frac{(x^2 + 8)-9}{(x + 1)(\sqrt{x^2 + 8}+3)}$$
Simplify the numerator:
$$\lim_{x \to -1} \frac{x^2 - 1}{(x + 1)(\sqrt{x^2 + 8}+3)}$$
Factor the numerator as a difference of squares:
$$\lim_{x \to -1} \frac{(x - 1)(x + 1)}{(x + 1)(\sqrt{x^2 + 8}+3)}$$

Step2: Cancel common factors

Cancel out the common factor of $(x + 1)$ in the numerator and the denominator (note that $x\to - 1$ but $x
eq - 1$ at the limit point, so cancellation is valid):
$$\lim_{x \to -1} \frac{x - 1}{\sqrt{x^2 + 8}+3}$$

Step3: Substitute $x=-1$

Now we can substitute $x = - 1$ into the simplified expression:
$$\frac{-1 - 1}{\sqrt{(-1)^2+8}+3}=\frac{-2}{\sqrt{1 + 8}+3}=\frac{-2}{3 + 3}=\frac{-2}{6}=-\frac{1}{3}$$

Answer:

$-\dfrac{1}{3}$