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The limit expression is:

\[
\lim_{t \to 36} \frac{\sin(\sqrt{t} - 6)}{t - 36}
\]

This is an indeterminate form of type \(\frac{0}{0}\) since:

- As \(t \to 36\), \(\sqrt{t} - 6 \to 0\), and hence \(\sin(\sqrt{t} - 6) \to 0\).
- The denominator \(t - 36 \to 0\).

To solve it, we can apply **L'Hôpital's Rule**, which states that for indeterminate forms like \(\frac{0}{0}\), we can take the derivative of the numerator and the denominator separately and then evaluate the limit.

### Step 1: Differentiate the numerator and denominator
- Numerator: \(\frac{d}{dt}[\sin(\sqrt{t} - 6)] = \cos(\sqrt{t} - 6) \cdot \frac{d}{dt}[\sqrt{t} - 6] = \cos(\sqrt{t} - 6) \cdot \frac{1}{2\sqrt{t}}\)
- Denominator: \(\frac{d}{dt}[t - 36] = 1\)

### Step 2: Apply L'Hôpital's Rule
The limit becomes:

\[
\lim_{t \to 36} \frac{\cos(\sqrt{t} - 6) \cdot \frac{1}{2\sqrt{t}}}{1}
\]

### Step 3: Evaluate the limit
Substitute \(t = 36\):

\[
\cos(\sqrt{36} - 6) \cdot \frac{1}{2\sqrt{36}} = \cos(6 - 6) \cdot \frac{1}{2 \times 6} = \cos(0) \cdot \frac{1}{12}
\]

Since \(\cos(0) = 1\), the result is:

\[
\frac{1}{12}
\]

Thus, the limit is:

\[
\lim_{t \to 36} \frac{\sin(\sqrt{t} - 6)}{t - 36} = \frac{1}{12}
\]

---

Would you like further clarification on any step? Here are some related questions to deepen your understanding:

1. What other conditions allow the use of L'Hôpital's Rule?
2. Can this limit be solved using a Taylor series expansion for small \((\sqrt{t} - 6)\)?
3. How would the approach change if the limit involved \(\cos(\sqrt{t} - 6)\) instead?
4. What is the geometric interpretation of this limit?
5. Could this method work for limits approaching infinity? Why or why not?

**Tip:** Always check the form of the limit before applying L'Hôpital's Rule to ensure it is indeterminate (like \(0/0\) or \(\infty/\infty\)).

Evaluate the limit: lim(t → 36) (sin(√t - 6) / (t - 36))

Learn how to evaluate the limit lim(t → 36) (sin(√t - 6) / (t - 36)) using L'Hôpital's Rule. This detailed solution covers the step-by-step differentiation of the numerator and denominator to resolve the indeterminate form 0/0. Suitable for Grades 11-12 calculus students.

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