Use l'hopital's Rule to evaluate lim x in 0 of
(4x(cos 6x-1 )) / sin 3x-3x
That would be
4(cos(6x-1) - 6xsin(6x-1))
------------------------------
3cos(3x) - 3
at x=0 that is
4(cos(-1)-0)
----------------
3-3
-> -∞
To use L'Hôpital's Rule, we need to compute the derivatives of the numerator and denominator and evaluate the limit of the ratio as x approaches 0.
Let's start by finding the derivatives:
Numerator:
f(x) = 4x(cos(6x) - 1)
To find the derivative of this function, we need to use the product rule. Let u = 4x and v = cos(6x) - 1.
f'(x) = u'v + uv'
= (4)(cos(6x) - 1) + (4x)(-6sin(6x))
= 4cos(6x) - 4 + (-24xsin(6x))
Denominator:
g(x) = sin(3x) - 3x
To find the derivative of this function, we can use the original form of L'Hôpital's Rule, which states that if lim x→c f(x)/g(x) is of the indeterminate form 0/0 or ∞/∞, then lim x→c f'(x)/g'(x) can be computed instead.
g'(x) = 3cos(3x) - 3
Now, let's apply L'Hôpital's Rule:
lim x→0 (4cos(6x) - 4 + (-24xsin(6x))) / (3cos(3x) - 3)
Since this is still of the indeterminate form 0/0, we can apply L'Hôpital's Rule again:
Differentiating the numerator and denominator once more:
Numerator:
f''(x) = -144cos(6x) - 576xsin(6x)
Denominator:
g''(x) = -9sin(3x)
Now, let's reapply L'Hôpital's Rule:
lim x→0 (-144cos(6x) - 576xsin(6x))) / (-9sin(3x))
At this point, we can simplify the expression and substitute x = 0:
lim x→0 (-144cos(6x) - 576xsin(6x))) / (-9sin(3x))
= (-144cos(6(0)) - 576(0)sin(6(0))) / (-9sin(3(0)))
= (-144cos(0) - 0sin(0)) / (-9sin(0))
= (-144(1) - 0(0)) / (-9(0))
= -144 / 0
Since the denominator is 0, we have an indeterminate form ∞/0. To evaluate this limit further, we need additional information.