Use L'Hoplital's rule to find the limit. Lim x->0 (3-3cos(x))/(sin(4x))
surely you can find the derivative of both top and bottom, no?
To use L'Hôpital's rule, we need to have an indeterminate form of either 0/0 or ∞/∞. Let's simplify the expression before applying the rule.
We are given the limit:
lim x->0 (3 - 3cos(x))/(sin(4x))
To apply L'Hôpital's rule, we need to take the derivative of the numerator and the derivative of the denominator separately.
Derivative of the numerator:
d/dx (3 - 3cos(x)) = 0 + 3sin(x) = 3sin(x)
Derivative of the denominator:
d/dx sin(4x) = 4cos(4x)
Now we can rewrite the expression using the derivatives:
lim x->0 (3sin(x))/(4cos(4x))
Since we were given a limit approaching x = 0, we can substitute x = 0 into the expression:
(3sin(0))/(4cos(4(0))) = 0/4cos(0) = 0/4 = 0
Therefore, the limit lim x->0 (3 - 3cos(x))/(sin(4x)) equals 0 using L'Hôpital's rule.