https://file.aiquickdraw.com/mathbot/file/17264643361174wunaexb.jpg

The limit presented in the image is:

\[
\lim_{{x \to 0}} \frac{\sqrt{3x + 49} - 7}{x}
\]

To solve this limit, we can use algebraic techniques like multiplying by the conjugate of the numerator to eliminate the square root. Here are the steps:

### Step 1: Multiply by the conjugate
Multiply both the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{3x + 49} + 7\):

\[
\frac{\sqrt{3x + 49} - 7}{x} \cdot \frac{\sqrt{3x + 49} + 7}{\sqrt{3x + 49} + 7} = \frac{\left(\sqrt{3x + 49} - 7\right)\left(\sqrt{3x + 49} + 7\right)}{x\left(\sqrt{3x + 49} + 7\right)}
\]

### Step 2: Simplify the numerator
The numerator becomes a difference of squares:

\[
(\sqrt{3x + 49})^2 - 7^2 = (3x + 49) - 49 = 3x
\]

Thus, the expression simplifies to:

\[
\frac{3x}{x(\sqrt{3x + 49} + 7)}
\]

### Step 3: Cancel \(x\)
We can cancel \(x\) from the numerator and denominator:

\[
\frac{3}{\sqrt{3x + 49} + 7}
\]

### Step 4: Evaluate the limit
Now, take the limit as \(x \to 0\):

\[
\lim_{{x \to 0}} \frac{3}{\sqrt{3x + 49} + 7} = \frac{3}{\sqrt{3(0) + 49} + 7} = \frac{3}{\sqrt{49} + 7} = \frac{3}{7 + 7} = \frac{3}{14}
\]

### Final Answer:
The value of the limit is \(\frac{3}{14}\).

Would you like more details on any step?

Here are some related questions for further exploration:
1. How does multiplying by the conjugate help simplify expressions?
2. Can limits involving square roots always be solved using conjugates?
3. What is the general form of a difference of squares?
4. What other techniques can be used to evaluate limits at indeterminate forms?
5. How does this limit compare to L'Hopital's rule for solving indeterminate forms?

**Tip:** Whenever you encounter square roots in limits, using conjugates to eliminate the square root is often an effective strategy!

Evaluate the limit: lim (x → 0) (sqrt(3x + 49) - 7) / x

Learn how to evaluate the limit of (sqrt(3x + 49) - 7) / x as x approaches 0, using algebraic manipulation and the conjugate method to simplify square roots. Step-by-step explanation with detailed reasoning.

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