We are asked to evaluate the following limit using the Squeeze Theorem:

$\lim_{x \to 4} \left( (x - 4) \sin \left( \frac{4\pi}{7(x - 4)} \right) \right)$

Step-by-step Solution:

Let’s rewrite the limit by introducing a substitution to simplify the expression.

Define:

$y = x - 4$

Thus, as $x \to 4$, we have $y \to 0$. Substituting this into the original limit, we get:

$\lim_{y \to 0} \left( y \sin \left( \frac{4\pi}{7y} \right) \right)$

Now, focus on the behavior of $\sin \left( \frac{4\pi}{7y} \right)$ as $y \to 0$.

Using the Squeeze Theorem:

We know from the standard trigonometric limits that:

$\lim_{y \to 0} \left( \frac{\sin \left( \frac{4\pi}{7y} \right)}{\frac{4\pi}{7y}} \right) = 1$

So, we can rewrite the limit as:

$\lim_{y \to 0} y \cdot \sin \left( \frac{4\pi}{7y} \right) = \lim_{y \to 0} \frac{y}{\frac{7y}{4\pi}} = 0$

This follows because $\sin\left(\frac{4\pi}{7y}\right)$ is bounded by $-1$ and $1$, and the term $y$ approaches zero, making the entire expression tend to zero.

Thus, the limit is:

$\boxed{0}$

Would you like more details or have any other questions?

5 Related Questions:

1. What is the Squeeze Theorem, and how is it applied in limits?
2. How can we handle limits that involve trigonometric functions?
3. What are the conditions for using trigonometric limit identities like $\lim_{x \to 0} \frac{\sin x}{x} = 1$?
4. Can you explain other substitution techniques for solving limits?
5. How does the behavior of $\sin(x)$ affect the evaluation of limits near zero?

Tip:

For limits involving oscillatory functions like sine or cosine, bounding the function and using the Squeeze Theorem is a powerful method to simplify the problem.