The limit you're asking for is:

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

To solve this, you can manipulate the expression using the standard trigonometric limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$:

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

Since $5x$ approaches 0 as $x$ approaches 0, we can use the standard limit:

$\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1$

So, the expression simplifies to:

$\lim_{x \to 0} 5 \cdot 1 = 5$

Thus, the limit is $5$.

Would you like further clarification, or do you have any questions?

5 Related Questions:

1. Why does $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$?
2. How can trigonometric identities help in solving limits?
3. What is the significance of using substitution in solving limits?
4. How would you solve $\lim_{x \to 0} \frac{\tan(x)}{x}$?
5. What happens to $\frac{\sin(5x)}{x}$ as $x$ approaches infinity?

Tip:

When working with limits involving trigonometric functions, especially as the variable approaches 0, substituting the angle with a simple multiple of $x$ can simplify the process by leveraging known limits like $\frac{\sin(x)}{x}$.