0/0 is an indeterminate form. In order to find the limit of the given function, we can use L'Hôpital's Rule.
Let f(x) = 1 - cos(x) and g(x) = x*sin(x). We are looking for lim(x approaches 0) f(x)/g(x).
Taking the derivative of f(x) with respect to x, we get f'(x) = sin(x).
Taking the derivative of g(x) with respect to x, we get g'(x) = sin(x) + x*cos(x).
So, applying L'Hôpital's Rule, we have:lim(x approaches 0) f(x)/g(x) = lim(x approaches 0) sin(x)/(sin(x) + x*cos(x))
Now, plugging in x = 0, we get 0/(0 + 0) = 0.
Therefore, lim(1-cos(x))/(x*sin(x)) as x approaches 0 is equal to 0.
0/0 is an indeterminate form. In order to find the limit of the given function, we can use L'Hôpital's Rule.
Let f(x) = 1 - cos(x) and g(x) = x*sin(x). We are looking for lim(x approaches 0) f(x)/g(x).
Taking the derivative of f(x) with respect to x, we get f'(x) = sin(x).
Taking the derivative of g(x) with respect to x, we get g'(x) = sin(x) + x*cos(x).
So, applying L'Hôpital's Rule, we have:
lim(x approaches 0) f(x)/g(x) = lim(x approaches 0) sin(x)/(sin(x) + x*cos(x))
Now, plugging in x = 0, we get 0/(0 + 0) = 0.
Therefore, lim(1-cos(x))/(x*sin(x)) as x approaches 0 is equal to 0.