Finding limit of trig equation(not sure if I should differentiate)

1. May 28, 2012

rectifryer

1. The problem statement, all variables and given/known data

Lim (cosθ-√3/2)/(θ-$pi$/6)
θ→$pi$/6
2. Relevant equations

3. The attempt at a solution

My attempt at this has been to try to multiply both the numerator and denominator by either the numerator's or denominator's conjugate. both result in 0 at the denominator.

I also factored out θ from the numerator and denominator but that still results in a 0 for the denominator.

Am I supposed to be able to differentiate this? It doesn't seem possible. How am I supposed to solve this?

Thanks in advance guys. Sorry if this is in the wrong format.

2. May 28, 2012

rectifryer

If I use L'Hopital's rule then it comes out with the right answer, however, my teacher hasn't covered this rule and I don't think we are supposed to use it.

3. May 28, 2012

sharks

Your limit appears to be:
$$\lim_{\theta \to \pi/6} \frac{\cos \theta - \sqrt{3}/2}{\theta - \pi /6}=\frac{0}{0}$$which is an indeterminate form. You have to use L'Hopital's rule to evaluate this limit. If your teacher hasn't covered this yet, then you should give the answer in its indeterminate form.

4. May 28, 2012

rectifryer

Thanks for writing that out in latex. I have bookmarked your sig haha.

I will email my teacher. We are doing problems in MathXL and there seems to be some disconnects from what it asks and what the teacher appears to have asked.

5. May 28, 2012

Bohrok

It's very possible to do this without l'Hôpital's rule. If you have a trig limit like this where the denominator is something like θ - π/6, try a substitution like x = θ - π/6 to make the denominator just a variable, and you also change the value used under the limit.
$$\text{Let }x = \theta - \frac{\pi}{6} \Longrightarrow \theta = x + \frac{\pi}{6}$$$$\text{As } \theta\to\frac{\pi}{6}, x\to 0$$$$\lim_{\theta \to \pi/6} \frac{\cos \theta - \sqrt{3}/2}{\theta - \pi /6} = \lim_{x\to 0}\frac{\cos(x + \pi/6) - \sqrt{3}/2}{x}$$Expand cos(x + π/6) and use the limits for (cosx - 1)/x and sinx/x as x→0 when working out the new limit expression.

6. May 28, 2012

SammyS

Staff Emeritus
You do not have to use L'Hôpital's rule to evaluate this.

However, you may use L'Hôpital's rule (if only you teacher would allow it.)

7. May 28, 2012

rectifryer

Incredible. Thanks a lot everyone!

8. May 28, 2012

algebrat

It looks like the definition of derivative of cos at pi over 6. So negative sin at pi over 6, which is -1/2

9. May 28, 2012

SammyS

Staff Emeritus
Good point !