Lim [(x)*cos(x) - sin(x)]/x^3 when x-->0 =?

Answer 1

Both numerator and denominator are defined at x = 0:

0 * cos(0) - sin(0) = 0 * 1 - 0 = 0.
0^3 = 0.

What we have is a limit of the indeterminate form 0/0. We can use L'hopital's rule to evaluate the limit.

The derivative of the numerator is cos(x) - x*sin(x) - cos(x)
= -x*sin(x).

The derivative of the denominator is 3x^2.

When x = 0, -[0*sin(0)] = 0, and 3*0^2 = 0.

It's still indeterminate, so let's use L'hopital's Rule again.

The derivative of -x*sin(x) is -sin(x) - x*cos(x) .

The derivative of 3x^2 is 6x.

When x = 0, -sin(0) - 0*cos(0) = 0 - 0*1 = 0, and 6(0) = 0.

Once again, it's indeterminate. Use L'hopital's rule yet again.

The derivative of -sin(x) - x*cos(x) is -cos(x) - cos(x) + x*sin(x)
= -2*cos(x) + x*sin(x)

The derivative of 6x is 6.

When x = 0, -2*cos(0) + 0*sin(0) = -2*1 + 0*0 = -2, and 6...well, it equals 6.

The form is no longer indeterminate. We have -2/6 = -1/3, which should be the limit of the orignial ratio.

A quick check with my calculator shows that when x = 0.1 (in radians, because that's the unit we use in calculus), the original expression evaluates as -0.333000119

x = 0.01 gives -0.333330000.

As we approach 0, the value is approaching -1/3!

Answer 2

lim[x-->0}{xcos(x) - sin(x)}/(x^3) =
lim[xcos(x)/x^3] - lim[sin(x)/x^3}
= lim[cos(x)/x^2] - lim[sin(x)/x^3]
=ininity - zero

Answer 3

sweet