The problem involves finding the limit of the expression \(\lim_{x\rightarrow\frac{\pi}{6}}\frac{2\sin(x)-1}{6x-\pi}\), which falls under the subject area of calculus, specifically limits and trigonometric functions.
The discussion is ongoing with various approaches being explored. Some participants have provided guidance on expressing terms in the limit, while others are questioning how to manipulate the expressions to resolve the indeterminate form. There is no explicit consensus on a single method, but productive dialogue is occurring.
Participants note constraints such as the inability to use L'Hopital's Rule and the recurring \(0/0\) form in their attempts. There is also a focus on understanding the relationships between trigonometric identities and limits.
mtayab1994 said:because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.
Dick said:Are you saying that you don't see why it is equal to cos(u)?
mtayab1994 said:Is my answer of 1/(2√3) correct?
Dick said:Yes, it is. That's the same thing l'Hopital would give you. You can always use l'Hopital to check these even if you can't use it to solve the problem.