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.