What is  \(\lim_{x \to 0} \frac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4}\) equal to?

Calculation:

Given,

The function is \( f(x) = \sqrt{x^2 + 9} - 3 \) and \( g(x) = \sqrt{x^2 + 16} - 4 \).

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

Multiply both the numerator and denominator by their respective conjugates:

\( \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4} \)

Simplify the numerator:

\( (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2 \)

Simplify the denominator:

\( (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2 \)

Now, the expression becomes:

\( \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \)

Simplify and evaluate the limit:

\( \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \) becomes:

\( \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)

Hence, the correct answer is Option 3.