The limit you're asking for is:

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

To solve this, we can use the standard limit property:

$\lim_{x \to 0} \frac{\sin(ax)}{x} = a$

In your case, $a = 5$. Therefore:

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

So, the limit is $5$.

Would you like more details on this solution, or do you have any questions?

5 Related Questions:

1. How is the limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ derived?
2. What are some common trigonometric limits to know for calculus?
3. How does L'Hôpital's Rule apply to limits involving trigonometric functions?
4. What is the significance of the small angle approximation in trigonometry?
5. How would you evaluate $\lim_{x \to 0} \frac{\sin^2(x)}{x^2}$?

Tip:

When dealing with trigonometric limits, remember that limits involving $\sin(x)/x$ as $x$ approaches 0 are fundamental and often serve as a starting point for more complex limit problems.