Solving this limit without L'Hopital

Question

I'm trying to find the solution for the following limit without using L'Hopitals rule.

The indeterminate form of $\frac{0}{0}$ is obtained but both the conjugate and or squeeze theorem can't be applied here (I think). I know that the solution is supposed to be 3 but I can't see how to reach it.

$\lim \limits_{x \to 0} \frac{sin3x}{x}$

$\frac{\sin 3x}x=3\cdot\frac{\sin 3x}{3x}$, and you should know $\lim_{x\to 0}\frac{\sin 3x}{3x}$. — Brian M. Scott, Oct 27 '16 at 20:00

score 2 · Accepted Answer · answered Oct 27 '16 at 20:01

2

$$\lim \limits_{x \to 0} \frac{\sin3x}{x}=\lim \limits_{x \to 0} 3\frac{\sin3x}{3x}$$

answered Oct 27 '16 at 20:01

E.H.E

23,280

Does this rule have a name? I don't understand why this is allowed. – TheAlPaca02 Oct 27 '16 at 20:03
1

$\lim \limits_{x \to 0} \frac{\sin ax}{ax}=1$ – E.H.E Oct 27 '16 at 20:04

Solving this limit without L'Hopital

1 Answers1