trigonometric limit without l'hopital: (2sin(x)-1)/(cos3x)

[Blackness]

Hello, I'd really appreciate a push in the right direction:

Limit(where x goes to pi/6) of (2sin(x)-1)/(cos3x)

I tried converting the '1' to sin(3x), as well as cos(3x) to (sin(4x)-sin(2x))/2 after multiplying the expression by sin(x)/sin(x), but I'm not getting anywhere.

 

[skaa]

\(\displaystyle \cos(3x)=4\cos^3x-3\cos x=\cos x(4\cos^2x-3)=\cos x(4-4\sin^2x-3)=\cos x(1-4\sin^2x)=\cos x(1-2\sin x)(1+2\sin x)\)

Your limit is

\(\displaystyle \displaystyle \lim_{x\to\frac{\pi}{6}}\, \dfrac{2\sin(x)\, -\, 1}{\cos(x)\, (1\, -\, 2\sin(x))\, (1\, +\, 2\sin(x))}\)

. . . . .\(\displaystyle \displaystyle =\, -\, \lim_{x\to\frac{\pi}{6}}\, \dfrac{1}{\cos(x)\, (1\,+\,2\sin(x))}\)

. . . . .\(\displaystyle \displaystyle =-\,\dfrac{1}{\left(\dfrac{\sqrt{\strut 3\,}}{2}\,(1\, +\,1)\right)}\, =\, -\,\dfrac{\sqrt{\strut 3\,}}{3}\)

 

[stapel]

Hello, I'd really appreciate a push in the right direction:

Limit(where x goes to pi/6) of (2sin(x)-1)/(cos3x)

I tried converting the '1' to sin(3x)

How is "1" equivalent to "sin(3x)" for all x?

as well as cos(3x) to (sin(4x)-sin(2x))/2

How?

I'm not getting anywhere.

Please reply showing your work, so we can try to figure out what's going on. Thank you!

 

[Blackness]

@skaa: Thank you very much!

@stapel: I get the message, I'll go into more detail next time i ask for help