To solve this limit:

$\lim_{h \to 0} \frac{\sin\left(\frac{7\pi}{6} + h\right) + \frac{1}{2}}{h}$

we can analyze the expression step-by-step.

Step 1: Understanding the Sine Expression

Since we know that $\frac{7\pi}{6}$ is in the third quadrant where sine is negative, let's calculate: $\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$

Thus, the expression inside the sine function can be rewritten as: $\sin\left(\frac{7\pi}{6} + h\right) = -\frac{1}{2} + h \cos\left(\frac{7\pi}{6}\right) + o(h)$

Where $o(h)$ denotes higher-order terms that tend to zero faster than $h$ as $h \to 0$.

Step 2: Substitute and Simplify the Expression

Now, we rewrite the original limit by substituting $\sin\left(\frac{7\pi}{6} + h\right)$ with its approximation: $\lim_{h \to 0} \frac{\left(-\frac{1}{2} + h \cos\left(\frac{7\pi}{6}\right)\right) + \frac{1}{2}}{h}$

Step 3: Simplify the Numerator

The terms $-\frac{1}{2}$ and $+\frac{1}{2}$ cancel each other, so we are left with: $\lim_{h \to 0} \frac{h \cos\left(\frac{7\pi}{6}\right)}{h}$

Now, divide by $h$: $\lim_{h \to 0} \cos\left(\frac{7\pi}{6}\right)$

Step 4: Calculate $\cos\left(\frac{7\pi}{6}\right)$

Since $\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, we get: $\lim_{h \to 0} \frac{\sin\left(\frac{7\pi}{6} + h\right) + \frac{1}{2}}{h} = -\frac{\sqrt{3}}{2}$

Final Answer

$\boxed{-\frac{\sqrt{3}}{2}}$

Would you like further details or explanations on any part of this solution?

Further Questions for Exploration

1. How can this limit be evaluated using L'Hôpital's Rule?
2. What are other methods to evaluate limits involving trigonometric functions?
3. Why is $\cos\left(\frac{7\pi}{6}\right)$ negative?
4. How does Taylor series expansion help in solving limits?
5. What are the differences between limits involving small-angle approximations and exact values?

Tip

For trigonometric limits involving small angles, approximating functions using Taylor series can simplify calculations.