The limit problem presented involves evaluating lim_{x→π/6} (2sin(x) - 1) / (6x - π). The substitution method initially leads to an indeterminate form of 0/0. The discussion highlights the effectiveness of L'Hôpital's Rule, yielding a result of cos(x)/3 at x=π/6, which simplifies to 1/(2√3). Ultimately, the correct limit is confirmed as 1/(2√3), aligning with the result obtained through L'Hôpital's Rule.
Students studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators looking for examples of limit evaluation techniques.
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.