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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?

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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!

Evaluate the limit: lim(h→0) (√(h+7) - 7)/h. If the limit doesn't exist enter F.

Learn how to evaluate the limit of (√(h+7) - 7)/h as h approaches 0 using the conjugate method. This step-by-step solution involves multiplying by the conjugate to simplify square roots, making it ideal for students in Grades 11-12.

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