Solving limit in many ways: lim[x->pi/6] [ (2cos(x) - sqrt{3}) / (6x - pi) ]

[tsunamari]

i want to know other ways to solve this limit other than using l'hôpital's rule or differentiable definition(f(x)-f(a)/x-a)

$\displaystyle \lim _{x\to \frac{\pi }{6}}\left(\frac{2\cos \left(x\right)-\sqrt{3}}{6x-\pi }\right)$

 

[Deleted member 4993]

i want to know other ways to solve this limit other than using l'hôpital's rule or differentiable definition(f(x)-f(a)/x-a)

$\displaystyle \lim _{x\to \frac{\pi }{6}}\left(\frac{2\cos \left(x\right)-\sqrt{3}}{6x-\pi }\right)$

Are you allowed to use Taylor series expansion of cos(pi/6 + h)?

 

[tsunamari]

Are you allowed to use Taylor series expansion of cos(pi/6 + h)?

hmm i dont think so is there any other way than that?

 

[mmm4444bot]

(f(x)-f(a))/(x-a)

A quick note about notation: missing grouping symbols shown in red. :cool:

 

[lookagain]

hmm i dont think so is there any other way than that?

Yes. I worked this out. You might approach it with the following method.


Hints:

Rewrite it as:

$\displaystyle \displaystyle\lim_{x \to \tfrac{\pi}{6}}\bigg(\dfrac{2cos(x) \ - \ \sqrt{3}}{6(x - \tfrac{\pi}{6})}\bigg)$


Let $\displaystyle \ y \ = \ x - \tfrac{\pi}{6}$


As x approaches pi/6, y approaches 0.


Substitute in the y expression:


$\displaystyle \displaystyle\lim_{y \to 0}\bigg(\dfrac{2cos(y + \tfrac{\pi}{6}) \ - \ \sqrt{3}}{6y}\bigg)$


Use cos(A + B), expand, and simplify.


And for this problem, assume that you have prior knowledge that


$\displaystyle \displaystyle\lim_{y \to 0}\dfrac{cos(y) - 1}{y} \ = \ 0 \ \ \ \ and \ \ \ \ \displaystyle\lim_{y \to 0}\dfrac{sin(y)}{y} \ = \ 1$


You must complete the supporting steps in between the hints, and following the last hints.