The problem shown in the image is the following limit:

$\lim_{x \to 0} \frac{\sin(5x)}{4x^2}.$

Solution:

We start by analyzing the expression. Notice that:

• The numerator is $\sin(5x)$, which approaches $0$ as $x \to 0$.
• The denominator $4x^2$ also approaches $0$ as $x \to 0$.

Thus, we have an indeterminate form $0/0$, so we can apply L'Hôpital's Rule, which states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate, then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.$

Step 1: Differentiate numerator and denominator

• The derivative of $\sin(5x)$ is $5\cos(5x)$.
• The derivative of $4x^2$ is $8x$.

Thus, the limit becomes:

$\lim_{x \to 0} \frac{5\cos(5x)}{8x}.$

Step 2: Apply L'Hôpital's Rule again

The new numerator is $5\cos(5x)$, which approaches $5$ as $x \to 0$. The new denominator $8x$ still approaches $0$, so we apply L'Hôpital's Rule again.

Differentiate numerator and denominator:

• The derivative of $5\cos(5x)$ is $-25\sin(5x)$.
• The derivative of $8x$ is $8$.

Thus, the limit becomes:

$\lim_{x \to 0} \frac{-25\sin(5x)}{8}.$

Step 3: Evaluate the limit

As $x \to 0$, $\sin(5x) \to 0$. Therefore, the entire expression becomes:

$\lim_{x \to 0} \frac{-25\sin(5x)}{8} = 0.$

Final Answer:

$\boxed{0}$

Do you want further details, or have any questions about the process?
Here are some related questions you can explore:

1. How is L'Hôpital's Rule derived, and when is it valid?
2. Can this problem be solved using Taylor expansions instead of L'Hôpital's Rule?
3. What happens if the denominator was $4x^3$ instead of $4x^2$?
4. How do trigonometric limits like $\lim_{x \to 0} \frac{\sin(kx)}{x}$ work?
5. What are some practical applications of limits in calculus?

Tip: Always check the form of a limit before applying L'Hôpital's Rule; indeterminate forms like $0/0$ and $\infty/\infty$ are key cases where it applies!