Evaluate the lim x^2-xsinx/x^2-sin^2x ?

To evaluate the limit of (x^2 - x*sin(x))/(x^2 - sin^2(x)) as x approaches 0, we can use L'Hôpital's rule.

Taking the derivative of the numerator and denominator separately, we have:

Numerator:
d/dx (x^2 - x*sin(x)) = 2x - (x*cos(x) + sin(x))

Denominator:
d/dx (x^2 - sin^2(x)) = 2x - 2*sin(x)*cos(x)

Now, we can evaluate the limit as x approaches 0:

lim(x->0) (2x - (x*cos(x) + sin(x))) / (2x - 2*sin(x)*cos(x))

Plugging in x = 0, we get:

(2*0 - (0*cos(0) + sin(0))) / (2*0 - 2*sin(0)*cos(0))
= (0 - (0 + 0)) / (0 - 2*0*1)
= 0 / 0

Since we have an indeterminate form of 0/0, we can apply L'Hôpital's rule again.

Taking the derivative of the numerator and denominator again, we have:

Numerator:
d/dx (2x - (x*cos(x) + sin(x))) = 2 - (cos(x) - x*sin(x))

Denominator:
d/dx (2x - 2*sin(x)*cos(x)) = 2 - 2*(cos^2(x) - sin^2(x))

Now, we can evaluate the limit as x approaches 0:

lim(x->0) (2 - (cos(x) - x*sin(x))) / (2 - 2*(cos^2(x) - sin^2(x)))

Plugging in x = 0, we get:

(2 - (cos(0) - 0*sin(0))) / (2 - 2*(cos^2(0) - sin^2(0)))
= (2 - (1 - 0)) / (2 - 2*(1 - 0))
= (2 - 1) / (2 - 2)
= 1 / 0

Again, we have an indeterminate form of 1/0. This suggests that the limit does not exist as x approaches 0.