To solve this limit:

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

we can use a standard limit identity:

$\lim_{{x \to 0}} \frac{\sin kx}{x} = k$

where $k$ is a constant. In this case, we have $k = 5$.

Solution Steps

1. Rewrite the expression: $\lim_{{x \to 0}} \frac{\sin 5x}{x} = \lim_{{x \to 0}} \left( \frac{\sin 5x}{5x} \cdot 5 \right)$

2. Separate the constants: $= 5 \cdot \lim_{{x \to 0}} \frac{\sin 5x}{5x}$

3. Use the identity $\lim_{{x \to 0}} \frac{\sin 5x}{5x} = 1$: $= 5 \cdot 1 = 5$

Final Answer

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

Would you like a detailed breakdown of each step or have any other questions?

Here are some related questions for further practice:

1. What is $\lim_{{x \to 0}} \frac{\sin 3x}{x}$?
2. How would you evaluate $\lim_{{x \to 0}} \frac{\tan 2x}{x}$?
3. Can you find $\lim_{{x \to 0}} \frac{\sin 7x}{2x}$?
4. How does the limit $\lim_{{x \to 0}} \frac{\sin kx}{kx} = 1$ help in simplifying trigonometric limits?
5. What happens if you evaluate $\lim_{{x \to 0}} \frac{\sin x}{x^2}$?

Tip: Recognizing standard limits, like $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$, helps in solving many trigonometric limit problems efficiently.