https://file.aiquickdraw.com/mathbot/file/1730236216821r78nb0xm.jpg

The expression in the image is:

\[
\lim_{{h \to 0}} \frac{\sqrt[3]{8+h}}{h}
\]

To solve this limit, let's analyze it:

1. **Direct Substitution**: If we directly substitute \( h = 0 \), we get:
   \[
   \frac{\sqrt[3]{8 + 0}}{0} = \frac{2}{0}
   \]
   which is undefined (division by zero). Therefore, direct substitution doesn’t work here, so we need a different approach.

2. **Rewrite and Expand**: Since this involves a cube root, we can use the approximation for small \( h \):
   \[
   \sqrt[3]{8 + h} \approx 2 + \frac{h}{12}
   \]
   where \( \sqrt[3]{8} = 2 \) and we use the linear approximation for small values of \( h \) around zero.

3. **Substitute the Approximation**:
   Substitute \( \sqrt[3]{8 + h} \approx 2 + \frac{h}{12} \) into the limit:
   \[
   \lim_{{h \to 0}} \frac{2 + \frac{h}{12} - 2}{h} = \lim_{{h \to 0}} \frac{\frac{h}{12}}{h} = \lim_{{h \to 0}} \frac{1}{12} = \frac{1}{12}
   \]

Thus, the answer is:
\[
\frac{1}{12}
\]

Would you like more details or have any questions on this solution?

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### Related Questions:
1. How can we apply different approximations for limits involving cube roots?
2. What is the general method for handling indeterminate forms like \( \frac{0}{0} \) in limits?
3. How does the cube root approximation compare to other methods like L'Hôpital's Rule?
4. Can we solve this limit using L'Hôpital's Rule, and what would be the process?
5. Why does direct substitution not work for most indeterminate limits?

### Tip:
For limits involving roots, try using Taylor expansions or approximations when direct substitution leads to indeterminate forms.

Explore the solution to the limit of (8 + h)^(1/3) / h as h approaches 0. This problem involves understanding limits, cube roots, and approximations to handle indeterminate forms effectively. Suitable for advanced high school students in Grades 11-12.

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