Solving a Limit Problem: Can't Grasp the Solution

[Tbarqawi]

So I've been trying to solve this limit problem for some time. Here is the problem:-
<br /> \lim_{x\rightarrow 0} {\frac{6sin(x) - 2sin(3x)}{tan^3(3x)}}<br />


I cannot use l'hopital's rule to solve it. I've tried taking 2 as a factor, then trying to use a trig identity, but I couldn't figure a thing. Dividing by x doesn't work either.

I have a feeling this problem is easy, yet I can't grasp the solution. So I figured I would get some help here.

 

[Ray Vickson]

So I've been trying to solve this limit problem for some time. Here is the problem:-
<br /> \lim_{x\rightarrow 0} {\frac{6sin(x) - 2sin(3x)}{tan^3(3x)}}<br />


I cannot use l'hopital's rule to solve it. I've tried taking 2 as a factor, then trying to use a trig identity, but I couldn't figure a thing. Dividing by x doesn't work either.

I have a feeling this problem is easy, yet I can't grasp the solution. So I figured I would get some help here.

Do you mean that you are not allowed to use l'Hospital's rule, or do you mean you don't know how to use it in this problem?

 

[Tbarqawi]

Yeah the textbook does not allow it.

 

[Ray Vickson]

Yeah the textbook does not allow it.

What ARE you allowed to use? If we don't know that we cannot make any sensible suggestions.

 

[Tbarqawi]

Ok, let me explain. The textbook has no mention of l'hopital's rule, thus we cannot use it. The way we are supposed to solve limits is by the "theorem" : the lim as x approaches 0 of sin(a*x) / sin(b*x) = a / b. To solve trigonometric limits, we use trigonometric identities usually to reach a state where we can use this theorem to "get rid" of what makes the denominator zero and then get the answer by substituting.

 

[Char. Limit]

Can you decompose tan(x) into sin(x)/cos(x) and work from there?