$$\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}$$
I know I can use L'Hospital's but I want to understand this particular explanation. They seem to skip something, and I'm not seeing the connection:
The limit is $\frac{6}{2}=3$ since $\lim_{x\to 0}\frac{\sin(x)}{x}=1$
Answer
For $x \neq 0$ and $x$ close to zero, we have
$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{\sin 2x} = 3 \cdot \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}$$
See what to do now?
No comments:
Post a Comment