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!