How to solve this limit

[juanpnav]

Without using l'hopital.

 

[BigBeachBanana]

Which function is [imath]sen[/imath]?

 

[topsquark]

Which function is [imath]sen[/imath]?

It's the Spanish version of sin(x).

-Dan

 

[MarkFL]

Have you tried L'Hopital?

 

[juanpnav]

Have you tried L'Hopital?

Yeah and it worked just fine. But I need to know a way to solve it without using L'hopital.

 

[MarkFL]

Yeah and it worked just fine. But I need to know a way to solve it without using L'hopital.

Oops, I obviously did not read your initial post well.

 

[pka]

Without using l'hopital.

View attachment 35188

Do you not know that [imath]\displaystyle\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin (\theta )}}{\theta } = 1~?[/imath]

 

[BigBeachBanana]

Without using l'hopital.

View attachment 35188

First, let [imath]u = x - \dfrac{\pi}{6} \implies \cos(x) = \cos\left(\dfrac{\pi}{6}+u\right) [/imath]
Use the identities below to turn all trig function into half angles i.e. [imath]\dfrac{u}{2}[/imath]

Trig Identities:
[imath] (1): \sin(2\theta) = 2\sin(\theta)\cos(\theta)\\ (2): \cos(A+B) =\cos(A)\cos(B)+\sin(A)\sin(B)\\ (3): 1-\cos(2\theta) = 2\sin^2(\theta) [/imath]

 

[juanpnav]

First, let [imath]u = x - \dfrac{\pi}{6} \implies \cos(x) = \cos\left(\dfrac{\pi}{6}+u\right) [/imath]
Use the identities below to turn all trig function into half angles i.e. [imath]\dfrac{u}{2}[/imath]

Trig Identities:
[imath] (1): \sin(2\theta) = 2\sin(\theta)\cos(\theta)\\ (2): \cos(A+B) =\cos(A)\cos(B)+\sin(A)\sin(B)\\ (3): 1-\cos(2\theta) = 2\sin^2(\theta) [/imath]

Thank you, could you expand please?

 

[BigBeachBanana]

Let [imath]u = x - \pi/6[/imath]

[math]\lim_{u \to 0} ~ \dfrac{\sin(u)}{ \dfrac{\sqrt{3}}{2}-\cos\left(\dfrac{\pi}{6}+u\right)}[/math]
Numerator:
[math]\sin(u) = 2\sin\left(\dfrac{u}{2}\right)\cos\left(\dfrac{u}{2}\right) \quad \because (1)[/math]
Denominator:
[math] \dfrac{\sqrt{3}}{2}-\cos\left(\dfrac{\pi}{6}+u\right) = \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\cos(u)+ \dfrac{1}{2}\sin(u)\quad \because (2) [/math]
Can you turn all trig functions in the denominator into half angles using the provided identities? Things will simplify and eventually you can just do the direct substitution. Post back your work if you need more help.

 

[juanpnav]

Ohh ofc, I forgot about sin cos of 30° is √3/2 and 1/2. Thank you, you life saver.