Question

Consider the following expression.
\[ \lim_{x \to -3} \frac{\sqrt{2x + 22} - 4}{x + 3} \] The value of the above expression (rounded to 2 decimal places) is \(\underline{\hspace{2cm}}\).

Hint:

For limits involving square roots, rationalizing the numerator helps remove indeterminate forms.

Answer 1

Correct Answer: 0.25

Substituting \( x = -3 \) directly gives an indeterminate form \( \frac{0}{0} \), so we rationalize the numerator.
\[ \frac{\sqrt{2x+22} - 4}{x+3} \times \frac{\sqrt{2x+22} + 4}{\sqrt{2x+22} + 4} \] \[ = \frac{(2x+22) - 16}{(x+3)(\sqrt{2x+22}+4)} \] \[ = \frac{2x + 6}{(x+3)(\sqrt{2x+22}+4)} \] \[ = \frac{2(x+3)}{(x+3)(\sqrt{2x+22}+4)} \] Canceling \( (x+3) \): \[ = \frac{2}{\sqrt{2x+22}+4} \] Now substitute \( x = -3 \): \[ = \frac{2}{\sqrt{16}+4} = \frac{2}{8} = 0.25 \] Final Answer: \[ \boxed{0.25} \]