To evaluate the limit as x approaches 64 of (sin((sqrt x)-8))/(x-64), we can use L'Hopital's rule, which states that if the limit of the ratio of two functions as x approaches a finite number is of the form 0/0 or ∞/∞, then we can differentiate the numerator and denominator and evaluate the limit of the ratio of the derivatives.
First, let's simplify the numerator:
sin((sqrt x) - 8) = sin(sqrt x) cos(-8) - cos(sqrt x) sin(-8)
= sin(sqrt x) cos 8 + cos(sqrt x) sin 8
= sin(sqrt x)
Next, let's differentiate the numerator and denominator:
d/dx (sin(sqrt x)) = (1/2) cos(sqrt x) / sqrt x
d/dx (x - 64) = 1
Now, let's evaluate the limit of the ratio of the derivatives:
lim x -> 64 (1/2) cos(sqrt x) / sqrt x / 1 = (1/2) cos(sqrt 64) / sqrt 64
= (1/2) cos(8) / 8
= (1/2) (1) / 8
= 1/16
Therefore, the limit as x approaches 64 of (sin((sqrt x)-8))/(x-64) is 1/16.
