To solve the given limit

$\lim_{x \to -7} \frac{(x + 7)^4 \sin\left(\frac{1}{(x + 7)^2}\right)}{\sin\left((x + 7)^2\right)},$

we can approach it by using some limit properties and approximations for small angles. Let's denote $y = x + 7$. As $x \to -7$, we have $y \to 0$. Thus, the limit becomes:

$\lim_{y \to 0} \frac{y^4 \sin\left(\frac{1}{y^2}\right)}{\sin(y^2)}.$

For small $y$, the approximations $\sin(z) \approx z$ can be useful. However, $\sin\left(\frac{1}{y^2}\right)$ oscillates rapidly and is bounded between -1 and 1. Therefore:

1. Consider the upper and lower bounds of $\sin\left(\frac{1}{y^2}\right)$:

$-1 \leq \sin\left(\frac{1}{y^2}\right) \leq 1.$

2. Now let's examine the effect of $y^4 \sin\left(\frac{1}{y^2}\right)$ for $y \to 0$:

Since $y^4$ approaches 0 as $y \to 0$, $y^4 \sin\left(\frac{1}{y^2}\right)$ also approaches 0, regardless of the oscillations of $\sin\left(\frac{1}{y^2}\right)$.

3. For the denominator $\sin(y^2)$:

As $y \to 0$,

$\sin(y^2) \approx y^2.$

4. Substitute these approximations back into the limit:

$\lim_{y \to 0} \frac{y^4 \sin\left(\frac{1}{y^2}\right)}{\sin(y^2)} \approx \lim_{y \to 0} \frac{y^4 \sin\left(\frac{1}{y^2}\right)}{y^2} = \lim_{y \to 0} y^2 \sin\left(\frac{1}{y^2}\right).$

5. Since $y^2 \sin\left(\frac{1}{y^2}\right)$ is bounded by $-y^2$ and $y^2$:

$-y^2 \leq y^2 \sin\left(\frac{1}{y^2}\right) \leq y^2.$

As $y \to 0$, $y^2 \to 0$. Thus, by the Squeeze Theorem:

$\lim_{y \to 0} y^2 \sin\left(\frac{1}{y^2}\right) = 0.$

Therefore, the limit is:

$\lim_{x \to -7} \frac{(x + 7)^4 \sin\left(\frac{1}{(x + 7)^2}\right)}{\sin\left((x + 7)^2\right)} = 0.$

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

Here are 5 related questions for further practice:

1. Evaluate $\lim_{x \to 0} \frac{x^3 \sin\left(\frac{1}{x}\right)}{\sin(x)}$.
2. Find $\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$.
3. Compute $\lim_{x \to 0} \frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x^2)}$.
4. Determine $\lim_{x \to 0} \frac{\sin(x^3)}{x^3}$.
5. Solve $\lim_{x \to \infty} \frac{\sin(x)}{x}$.

Tip: When dealing with trigonometric functions involving limits, it is often useful to employ small-angle approximations and the Squeeze Theorem to handle oscillatory behavior.