Lim sin(60-x)/2cosx-1 x approaches 60
Use L'Hopitals rule
Lim sin(60°-x)/(2cosx - 1)
= lim sin(π/3-x)/(2cosx - 1) , as x --> π/3
= lim cos(π/3 - x)/(-2sinx) as x---> π/3
= cos 0/(-2sin(π/3
= 1/(-2(√3/2)
= -√3 or appr -.5773
http://www.wolframalpha.com/input/?i=y+%3D+sin%28%CF%80%2F3-x%29%2F%282cosx-1%29+
notice the vertical asymptote at appr -.5
I WAS STRUGGLING WITH THIS PROBLEM SO BUT FROM NOW ONWARD I WILL BEAR THIS AT THE BACK OF MY MIND
To find the limit of the given function as x approaches 60, we can use the concept of L'Hôpital's Rule. This rule can be used to evaluate limits involving indeterminate forms such as 0/0 or ∞/∞.
Let's start by applying L'Hôpital's Rule. Differentiating both the numerator and denominator with respect to x:
lim x→60 [sin(60 - x)] / [2cos(x) - 1]
= lim x→60 [cos(60 - x)] / [-2sin(x)]
Next, we substitute x = 60 into this expression:
= [cos(0)] / [-2sin(60)]
= 1 / [-2 * (sqrt(3)/2)]
= -1 / √3
= -√3 / 3
Therefore, the limit of the given function as x approaches 60 is -√3 / 3.