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The given limit is:

\[
\lim_{x \to 1} \frac{\sqrt{x^2 + 15} - 4}{x - 1}
\]

### Solution using L'Hôpital's Rule:
Since directly substituting \(x = 1\) into the expression gives a \( \frac{0}{0} \) indeterminate form, we can apply L'Hôpital's Rule, which requires differentiating the numerator and denominator.

1. **Numerator**: \(\sqrt{x^2 + 15} - 4\)
   - The derivative of \(\sqrt{x^2 + 15}\) is: 
   \[
   \frac{d}{dx}\left(\sqrt{x^2 + 15}\right) = \frac{1}{2\sqrt{x^2 + 15}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 15}}
   \]
   - The derivative of the constant \(-4\) is \(0\).

   So, the derivative of the numerator is:
   \[
   \frac{x}{\sqrt{x^2 + 15}}
   \]

2. **Denominator**: \(x - 1\)
   - The derivative of \(x - 1\) is just \(1\).

Using L'Hôpital's Rule, the limit becomes:
\[
\lim_{x \to 1} \frac{\frac{x}{\sqrt{x^2 + 15}}}{1}
\]
Substitute \(x = 1\):
\[
\frac{1}{\sqrt{1^2 + 15}} = \frac{1}{\sqrt{16}} = \frac{1}{4}
\]

Thus, the solution to the limit is:

\[
\boxed{\frac{1}{4}}
\]

Would you like further details or have any questions?

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Here are 5 related questions:
1. What are the conditions to apply L'Hôpital's Rule?
2. How would you solve this limit without using L'Hôpital's Rule?
3. What other indeterminate forms can L'Hôpital's Rule be used for?
4. Can we apply factorization techniques to solve this limit?
5. How does the chain rule apply in the differentiation of the numerator?

**Tip:** Always check for an indeterminate form before applying L'Hôpital's Rule.

lim_{x→1} (√(x² + 15) - 4) / (x - 1)

Learn how to solve the limit lim_{x→1} (√(x² + 15) - 4) / (x - 1) using L'Hôpital's Rule. This example shows step-by-step differentiation and application of the limit for students in Grades 11-13.

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