Evaluate the following limit
lim x -> 0 sin x[cos (1/x^2) + sin (1/x^2)]

The limit would present itself in the form

Obviously, you can't evaluate the two expressions

But since sine and cosine functions are always between -1 and 1, the sum of these will be a number between -2 and 2.

So, the limit presents itself in the form

Since you have the product of a quantity which tends to zero, sin(x), and a quantity which is bounded, cos(1/x^2)+sin(1/x^2).

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)