To find the limit of the function [tex]\(\lim _{x \rightarrow 4} \frac{x^2-16}{\sqrt{3 x+4}-4}\)[/tex], we need to carefully analyze and simplify the expression. Let’s go through the steps.
1. Identify the form of the function:
We begin by substituting [tex]\(x = 4\)[/tex] into the function to check if it forms an indeterminate form:
[tex]\[
\frac{4^2 - 16}{\sqrt{3 \cdot 4 + 4} - 4} = \frac{16 - 16}{\sqrt{12 + 4} - 4} = \frac{0}{\sqrt{16} - 4} = \frac{0}{4 - 4} = \frac{0}{0}
\][/tex]
The expression forms an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
2. Simplify the numerator using factoring:
Notice that the numerator [tex]\(x^2 - 16\)[/tex] is a difference of squares, which can be factored:
[tex]\[
x^2 - 16 = (x - 4)(x + 4)
\][/tex]
3. Simplify the denominator:
The denominator can be rewritten to recognize a favorable form:
[tex]\[
\sqrt{3x + 4} - 4
\][/tex]
4. Use rationalization:
To handle the square root in the denominator, multiply and divide by the conjugate of the denominator:
[tex]\[
\frac{(x - 4)(x + 4)}{\sqrt{3x + 4} - 4} \times \frac{\sqrt{3x + 4} + 4}{\sqrt{3x + 4} + 4} = \frac{(x - 4)(x + 4)(\sqrt{3x + 4} + 4)}{(\sqrt{3x + 4})^2 - 4^2}
\][/tex]
Simplify this expression. Note that the denominator becomes a difference of squares:
[tex]\[
(\sqrt{3x + 4})^2 - 4^2 = 3x + 4 - 16 = 3x - 12
\][/tex]
So we have:
[tex]\[
\frac{(x - 4)(x + 4)(\sqrt{3x + 4} + 4)}{3(x - 4)}
\][/tex]
5. Cancel common factors:
There is a common factor of [tex]\(x - 4\)[/tex] in both the numerator and the denominator:
[tex]\[
\frac{(x - 4)(x + 4)(\sqrt{3x + 4} + 4)}{3(x - 4)} = \frac{(x + 4)(\sqrt{3x + 4} + 4)}{3}
\][/tex]
For [tex]\(x \neq 4\)[/tex], this simplifies to:
[tex]\[
\frac{(x + 4)(\sqrt{3x + 4} + 4)}{3}
\][/tex]
6. Evaluate the limit:
Substitute [tex]\(x = 4\)[/tex] into the simplified expression:
[tex]\[
\frac{(4 + 4)(\sqrt{3 \cdot 4 + 4} + 4)}{3} = \frac{8(\sqrt{12 + 4} + 4)}{3} = \frac{8(\sqrt{16} + 4)}{3} = \frac{8(4 + 4)}{3} = \frac{8 \cdot 8}{3} = \frac{64}{3}
\][/tex]
So the final answer is:
[tex]\[
\lim _{x \rightarrow 4} \frac{x^2-16}{\sqrt{3 x+4}-4} = \frac{64}{3}
\][/tex]
