As I solve this problem, it is undefined. Can you explain how you get 3 or 0. One of this is the answer but I'm not sure.
lim as y approaches 0
6x^2 (cot y) (csc2y):
A. 3
B. 0
ctn y = siny/cosy
csc2 y= 1/sin2 y=1/(2sinycosy)
so ctn y * csc^2 y= 1/2cos^2 y
lim y>>0 6x^2/2cos^2y=3x^2
Did you type this correctly?
I meant 6y^2. Now I see where I went wrong.
To evaluate the limit as y approaches 0 of the given expression, we can begin by simplifying the expression using trigonometric identities.
Starting with the given expression: 6x^2 * cot(y) * csc(2y)
Now, cot(y) is the reciprocal of tan(y), which can be expressed as cos(y) divided by sin(y). Therefore, cot(y) = cos(y) / sin(y).
Similarly, csc(2y) is the reciprocal of sin(2y), which can be expressed as 1 divided by sin(2y).
Substituting these values into the expression, we have:
6x^2 * (cos(y) / sin(y)) * (1 / sin(2y))
Next, we can simplify this expression further. The sin(2y) in the denominator can be expanded as follows:
sin(2y) = sin(y + y) = sin(y)cos(y) + cos(y)sin(y) = 2sin(y)cos(y)
Now, substituting this value back into the expression, we have:
6x^2 * (cos(y) / sin(y)) * (1 / (2sin(y)cos(y)))
Canceling out the sin(y) term in the numerator and denominator, we get:
6x^2 * (cos(y) / (2cos(y))) = 3x^2
Finally, as y approaches 0, the value of 3x^2 remains unchanged. Therefore, the limit is equal to 3x^2.
So, the answer is A. 3.