The given expression is:

$\lim_{{x \to 0}} \frac{{\sqrt{49 - x} - 7}}{x}$

Step-by-step solution:

This is an indeterminate form of $\frac{0}{0}$ when directly substituting $x = 0$. So, we need to manipulate the expression to resolve this.

1. Multiply by the conjugate:

We can multiply the numerator and denominator by the conjugate of the numerator to simplify the square root term. The conjugate of $\sqrt{49 - x} - 7$ is $\sqrt{49 - x} + 7$.

$\lim_{{x \to 0}} \frac{{\sqrt{49 - x} - 7}}{x} \times \frac{{\sqrt{49 - x} + 7}}{{\sqrt{49 - x} + 7}} = \lim_{{x \to 0}} \frac{{(\sqrt{49 - x} - 7)(\sqrt{49 - x} + 7)}}{{x(\sqrt{49 - x} + 7)}}$

2. Simplify the numerator:

By applying the difference of squares formula $(a - b)(a + b) = a^2 - b^2$, we simplify the numerator:

$(\sqrt{49 - x})^2 - 7^2 = (49 - x) - 49 = -x$

Thus, the expression becomes:

$\lim_{{x \to 0}} \frac{{-x}}{{x(\sqrt{49 - x} + 7)}} = \lim_{{x \to 0}} \frac{{-1}}{{\sqrt{49 - x} + 7}}$

3. Evaluate the limit:

Now, substitute $x = 0$ into the expression:

$\frac{{-1}}{{\sqrt{49 - 0} + 7}} = \frac{{-1}}{{7 + 7}} = \frac{{-1}}{{14}}$

Final Answer:

$\lim_{{x \to 0}} \frac{{\sqrt{49 - x} - 7}}{x} = \frac{{-1}}{{14}}$

Would you like further explanation on any step or more examples of limits?

Here are 5 related questions for further practice:

1. How would you approach solving $\lim_{{x \to 0}} \frac{{\sqrt{x+1} - 1}}{x}$?
2. What is the limit of $\lim_{{x \to 0}} \frac{{1 - \cos(x)}}{x^2}$?
3. Can you compute $\lim_{{x \to 0}} \frac{{\sin(3x)}}{x}$?
4. How do we apply L'Hopital's Rule to solve limits of indeterminate forms?
5. What are the different techniques for evaluating limits involving radicals?

Tip: Always check if a limit expression is in an indeterminate form before applying algebraic manipulations or L'Hopital's Rule!