The problem presented in the image appears to involve evaluating the following limit:

$\lim_{x \to 0} \frac{\sin(5x)}{4x}.$

Solution:

Step 1: Simplify the Expression Using the Sine Limit Rule

We know that: $\lim_{u \to 0} \frac{\sin(u)}{u} = 1.$ To apply this rule, we rewrite the argument of $\sin(5x)$ in terms of $5x$. Factor $5$ out of the denominator:

$\frac{\sin(5x)}{4x} = \frac{\sin(5x)}{5x} \cdot \frac{5}{4}.$

Step 2: Apply the Limit

Now, we can split the limit into two parts:

$\lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5}{4}.$

From the sine limit rule: $\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1.$

Thus, the expression simplifies to: $1 \cdot \frac{5}{4} = \frac{5}{4}.$

Final Answer:

$\lim_{x \to 0} \frac{\sin(5x)}{4x} = \frac{5}{4}.$

Would you like me to provide more details or further examples?

Here are 5 related questions you may find interesting:

1. How does the limit $\lim_{x \to 0} \frac{\sin(kx)}{x}$ change if $k$ is replaced with another constant?
2. Can this result be derived using L'Hôpital's Rule?
3. How do we evaluate $\lim_{x \to 0} \frac{\tan(kx)}{x}$?
4. What are the conditions under which the small-angle approximation for sine is valid?
5. How does the unit circle interpretation help understand the sine function's behavior near zero?

Tip: When working with trigonometric limits, always check if a standard limit like $\frac{\sin(x)}{x} \to 1$ can be directly applied to simplify your work!