To evaluate the given limit:

$\lim_{{h \to 0}} \frac{\sqrt{h+7} - 7}{h}$

We will attempt to simplify this expression. To handle this, we use the technique of multiplying by the conjugate of the numerator to eliminate the square root. The conjugate of $\sqrt{h+7} - 7$ is $\sqrt{h+7} + 7$.

Step-by-step:

1. Multiply both the numerator and the denominator by the conjugate of the numerator: $\frac{\sqrt{h+7} - 7}{h} \times \frac{\sqrt{h+7} + 7}{\sqrt{h+7} + 7} = \frac{(\sqrt{h+7})^2 - 7^2}{h(\sqrt{h+7} + 7)}$

2. Simplify the numerator: $(\sqrt{h+7})^2 - 7^2 = (h+7) - 49 = h - 42$

   So, the expression becomes: $\frac{h - 42}{h(\sqrt{h+7} + 7)}$

3. Now, cancel $h$ from the numerator and denominator: $\frac{h - 42}{h(\sqrt{h+7} + 7)} = \frac{1}{\sqrt{h+7} + 7}$

4. Finally, take the limit as $h \to 0$: $\lim_{{h \to 0}} \frac{1}{\sqrt{h+7} + 7} = \frac{1}{\sqrt{0+7} + 7} = \frac{1}{\sqrt{7} + 7}$

Thus, the limit is:

$\frac{1}{\sqrt{7} + 7}$

Would you like more details or further clarification on this solution?

Here are some related questions:

1. How would the limit change if the function inside the square root was different, e.g., $\sqrt{h+9}$?
2. What are conjugates, and why are they useful in simplifying limits involving square roots?
3. How would you evaluate the limit if the expression inside the square root had a different form, such as $\sqrt{h+1}$?
4. Can L'Hopital's Rule be used for this limit, and if so, how?
5. How does multiplying by the conjugate simplify other types of limits involving radicals?

Tip: Always check if multiplying by the conjugate can simplify limits involving square roots—it’s a powerful technique!