Answer
$8$
Work Step by Step
We have to find $\lim\limits_{{x \to 3}} \frac{x^2 - 9}{\sqrt{x^2 + 7} - 4}$
But if we put $x = 3$ we get zero in both the numerator and the denominator.
To find the answer we can do the following steps:
Multiply the numerator and denominator by $\sqrt{x^2 + 7} + 4$
= $\frac{x^2 - 9}{\sqrt{x^2 + 7} - 4} \times \frac{\sqrt{x^2 + 7} + 4}{\sqrt{x^2 + 7} + 4}$
= $\frac{(x^2 - 9)(\sqrt{x^2 + 7} + 4)}{(\sqrt{x^2 + 7} - 4)(\sqrt{x^2 + 7} + 4)}$
Now, we know $(a - b)(a + b) = a^2 - b^2$
So, we get: $\frac{(x^2 - 9)(\sqrt{x^2 + 7} + 4)}{(\sqrt{x^2 + 7})^2 - 4^2}$
= $\frac{(x^2 - 9)(\sqrt{x^2 + 7} + 4)}{x^2 + 7 - 16}$
= $\frac{(x^2 - 9)(\sqrt{x^2 + 7} + 4)}{x^2 - 9}$
= $\sqrt{x^2 + 7} + 4$
Now, as $\frac{x^2 - 9}{\sqrt{x^2 + 7} - 4} $ = $\sqrt{x^2 + 7} + 4$
$\lim\limits_{{x \to 3}} \frac{x^2 - 9}{\sqrt{x^2 + 7} - 4} = \lim\limits_{{x \to 3}} (\sqrt{x^2 + 7} + 4)$
Now, we can directly substitute the value of $x = 3$.
Therefore, we get $\sqrt{(3)^2 + 7} + 4 = \sqrt{16} + 4 = 4 + 4 = 8$
Therefore,
$\lim\limits_{{x \to 3}} \frac{x^2 - 9}{\sqrt{x^2 + 7} - 4} = 8$
