L'Hopital Rule to evaluate this trig function

Question

I am supposed to use L'Hopital's rule to evaluate this question but do not know how.

lim 3x(cos(8x)-1) / sin(6x)-6x

x→0

Answer 1

In order to use L'Hopital's Rule, the limit must of the form 0/0, ∞/∞, or the like. In this case, plugging in x = 0 we see we have 0/0 so L'Hopital's Rule does apply. L'Hopital's Rule says the limit of f(x)/g(x) = limit f '(x)/g'(x)

f(x) = 3x(cos(8x) - 1)

Using the product rule, f '(x) = 3(cos(8x) - 1) + 3x(-8sin(8x))

f '(x) = 3cos(8x) - 3 - 24xsin(8x)

g(x) = sin(6x) - 6x

g'(x) = 6cos(6x) - 6

So the limit becomes:

Or, dividing top and bottom by 3:

lim (cos(8x) - 1 - 8xsin(8x))/(2cos(6x) - 2)

But that still results in 0/0 so, take the top and bottom derivatives again to get:

(-16sin(8x) - 64xcos(8x))/(-12sin(6x))

Still 0/0 so, take the top and bottom derivatives again to get:

lim (512xsin(8x) - 192cos(8x))/(-72cos(6x))

Now, when we plug in zero, we get -192/-72 = 8/3

Answer 2

You need to use Del' Hospital's rule twice. So you need to differentiate the numerator and the denominator two times.