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

[rectifryer]

Homework Statement

Lim (cosθ-√3/2)/(θ-pi/6)
θ→pi/6

Homework Equations

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.

 

[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.

 

[DryRun]

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.

 

[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.

 

[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.

 

[SammyS]

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.

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.)

Follow Bohrok's suggestion.

 

[rectifryer]

Incredible. Thanks a lot everyone!

 

[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

 

[SammyS]

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

Good point !