0
$\begingroup$

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.

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

$\endgroup$
  • 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.