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.

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

  • 1
    $\begingroup$ $\frac{\sin 3x}x=3\cdot\frac{\sin 3x}{3x}$, and you should know $\lim_{x\to 0}\frac{\sin 3x}{3x}$. $\endgroup$ – Brian M. Scott Oct 27 '16 at 20:00
  • $\begingroup$ Do you know equivalent functions? $\endgroup$ – Bernard Oct 27 '16 at 20:01

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

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  • $\begingroup$ Does this rule have a name? I don't understand why this is allowed. $\endgroup$ – TheAlPaca02 Oct 27 '16 at 20:03
  • 1
    $\begingroup$ $\lim \limits_{x \to 0} \frac{\sin ax}{ax}=1$ $\endgroup$ – E.H.E Oct 27 '16 at 20:04

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