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The given limit is:

\[
\lim_{h \to 0} \frac{\tan\left(\frac{\pi}{6} + h\right) - \tan\left(\frac{\pi}{6}\right)}{h}
\]

This is a standard derivative form, where the limit resembles the definition of the derivative of \(\tan(x)\) at \(x = \frac{\pi}{6}\).

We know the derivative of \(\tan(x)\) is \(\sec^2(x)\). So, the limit will be:

\[
\frac{d}{dx} \left[ \tan(x) \right] \text{ at } x = \frac{\pi}{6}
= \sec^2\left(\frac{\pi}{6}\right)
\]

Now, we calculate:

\[
\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}
\]

Thus:

\[
\sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}
\]

Therefore, the value of the limit is:

\[
\frac{4}{3}
\]

Let me know if you'd like any additional details or explanations!

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Here are 5 related questions to expand your understanding:

1. How do you find the derivative of trigonometric functions like sine and cosine?
2. What is the general process for evaluating limits that result in indeterminate forms?
3. How does the definition of the derivative apply to limits of this type?
4. What are some other examples of trigonometric limits involving the definition of the derivative?
5. How can you evaluate limits that do not involve standard derivatives?

**Tip:** When evaluating limits that resemble the derivative definition, identifying the function involved and calculating its derivative is a helpful shortcut.

Evaluate the limit: lim(h→0) (tan(π/6 + h) - tan(π/6)) / h

Learn how to evaluate the limit lim(h→0) (tan(π/6 + h) - tan(π/6)) / h using the definition of the derivative. This problem involves trigonometric functions and their derivatives, focusing on sec^2(x). Suitable for high school calculus students.

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