lim x->0 (x)sin(x)/1-cos(x).

They both go to 0 so 0/0.

now take L'H
and product rule of top?
lim x->0 (1)sin(x)+ cos(x)(x)/1-cos(x)

what next? how do i solve from here?

derivative of top / derivative of bottom

(x cos x + sin x) / (sin x)

(x cos x/sin x) + 1

x (1 -x^2/2 ....)/(x -x^3/3 ....) + 1

x/x + 1

2

or, using the rule again,

(cosx - x sinx + cosx)/cosx
= (2cosx - x sinx)/cosx
= (2-0)/1
= 2

To solve the given limit, you correctly recognized that it is in the form of 0/0, which is an indeterminate form. One way to evaluate such limits is to apply L'Hôpital's Rule.

First, let's differentiate the numerator and the denominator separately:

Numerator:
d/dx [(x)sin(x)] = sin(x) + x*cos(x)

Denominator:
d/dx [1-cos(x)] = sin(x)

Now, you can rewrite the limit using the derived numerator and denominator:

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

The next step is to simplify by canceling out the common factor of sin(x) in the numerator and denominator:

lim x->0 (1 + x*cos(x)/sin(x))

Next, you can evaluate the limit by substituting 0 into the expression:

lim x->0 (1 + 0*cos(0)/sin(0))

Since cos(0) equals 1 and sin(0) equals 0, we have:

lim x->0 (1 + 0/0)

Now, we encounter another indeterminate form, where we have 1 + 0/0. To resolve this, we can rewrite the expression as follows:

lim x->0 [(1/0)⋅(0/0)]

Now, using indeterminate form, we can find that 0/0 is 0:

lim x->0 [1/0]⋅0 = 0

Therefore, the value of the limit as x approaches 0 is 0.