trigonometric Limit calculation

[akleron]

Hello.
Trying to figure out how to calculate the following limit.
I got "0/0" kind of limit and we are not allowed to use l'hopital yet.
I thought the following trigonometric identity might help but I ended up with a wrong answer.
1st, I cant figure out where is my mistake (it is certainly there)
2nd, any suggestions for calculating this Limit ?


This is my best attempt (which is wrong)

 

[Dr.Peterson]

My first thought is to look for the cause of indeterminacy: at [MATH]x = \frac{4\pi}{3}[/MATH], [MATH]\tan^2 x - 3 = 0[/MATH] in the numerator, and [MATH]\cos\left(x - \frac{5\pi}{6}\right) = 0[/MATH] in the denominator.

I'm inclined to do what I suggested in another of your problems, namely using angle-sum identities to rewrite the offending tan and cos in term of [MATH]x - \frac{4\pi}{3}[/MATH], which approaches 0. Then you should be able to use familiar limits at 0 to finish.

On the other hand, no one in their right mind would use anything other than L'Hopital for such an ugly problem!

 

[Cubist]

OP, your working seems good (except you missed out the 23/22). In the denominator you did the right thing.

But in the numerator try this approach:- factor out a tan; then use difference of squares; then rewrite one tan (the -ve one) as sin/cos; then multiply numerator and denominator by cos. This is what I ended up with...

\(\displaystyle \frac{23\tan\left(x\right)\cdot \left(\sqrt{3}+\tan\left(x\right)\right)\left(\color{red} {\sqrt{3}\cdot \cos\left(x\right)-\sin\left(x\right)} \color{black} \right)}{11\cos\left(x\right)\cdot \left(\color{red} {\sqrt{3}\cdot \cos\left(x\right)-\sin\left(x\right)} \color{black} \right)} \)

 

[hoosie]

Once you get to the step:
simplify as follows:

By substituting back into original expression:

Hope these extra steps help in your understanding of the problem.

 

[Deleted member 4993]

OP, your working seems good (except you missed out the 23/22). In the denominator you did the right thing.

But in the numerator try this approach:- factor out a tan; then use difference of squares; then rewrite one tan (the -ve one) as sin/cos; then multiply numerator and denominator by cos. This is what I ended up with...

\(\displaystyle \frac{23\tan\left(x\right)\cdot \left(\sqrt{3}+\tan\left(x\right)\right)}{11\cos\left(x\right)} \)

I don't think we need to multiply by "1" after the reduction above. We can simply evaluate the "expression" at x = \(\displaystyle \frac{4\pi}{3}\) and get the answer (=\(\displaystyle -\frac{276}{11}\))

 

[Cubist]

I don't think we need to multiply by "1" after the reduction above. We can simply evaluate the "expression" at x = \(\displaystyle \frac{4\pi}{3}\) and get the answer (=\(\displaystyle -\frac{276}{11}\))

I wasn't very clear in my post, sorry. I did mean to say why I highlighted the red parts but I forgot!

After performing my suggested steps, the red factors will actually be present in the expression - and they are in fact the parts (in the numerator and denominator) that become 0 when \(\displaystyle x = \frac{4\pi}{3}\). In other words, the red parts were the stumbling block in the original expression. But in this form they can be nicely cancelled out and the limit can then be calculated.