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We are tasked with evaluating the limit:

\[
\lim_{x \to 4} \frac{3x - 12}{3 - \sqrt{2x + 1}}
\]

### Step 1: Substitution
First, substitute \(x = 4\) directly into the expression to check for any indeterminate form.

- For the numerator: \(3(4) - 12 = 12 - 12 = 0\)
- For the denominator: \(3 - \sqrt{2(4) + 1} = 3 - \sqrt{8 + 1} = 3 - \sqrt{9} = 3 - 3 = 0\)

Thus, we have a \(\frac{0}{0}\) indeterminate form, so we must apply further techniques such as algebraic manipulation or L'Hôpital's Rule.

### Step 2: Algebraic Simplification
The numerator can be factored as follows:

\[
3x - 12 = 3(x - 4)
\]

Now the limit becomes:

\[
\lim_{x \to 4} \frac{3(x - 4)}{3 - \sqrt{2x + 1}}
\]

### Step 3: Rationalizing the Denominator
To eliminate the square root, we can multiply the numerator and the denominator by the conjugate of the denominator:

\[
\frac{3 - \sqrt{2x + 1}}{3 - \sqrt{2x + 1}} \cdot \frac{3 + \sqrt{2x + 1}}{3 + \sqrt{2x + 1}} = \frac{3(x - 4)(3 + \sqrt{2x + 1})}{(3 - \sqrt{2x + 1})(3 + \sqrt{2x + 1})}
\]

The denominator simplifies using the difference of squares:

\[
(3 - \sqrt{2x + 1})(3 + \sqrt{2x + 1}) = 9 - (2x + 1) = 9 - 2x - 1 = 8 - 2x
\]

Thus, the limit becomes:

\[
\lim_{x \to 4} \frac{3(x - 4)(3 + \sqrt{2x + 1})}{8 - 2x}
\]

### Step 4: Simplifying and Applying the Limit
Now substitute \(x = 4\) into the remaining expression after canceling the factor \((x - 4)\) from the numerator and denominator:

The denominator \(8 - 2x\) becomes:

\[
8 - 2(4) = 8 - 8 = 0
\]

Thus, I will proceed further in solving the

Evaluate the limit: lim (x -> 4) (3x - 12) / (3 - sqrt(2x + 1))

Learn how to evaluate the limit lim (x -> 4) (3x - 12) / (3 - sqrt(2x + 1)) by using algebraic manipulation and rationalization. This problem involves applying L'Hôpital's Rule and simplifying square root expressions. Suitable for students in Grades 11-12.

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