i need to take the limit as x-->0 of 3x^2csc^2(x)...Is the answer 0?
Two things you need to know:
1. csc x = 1 / sin x
2. lim as x->0 (sin x / x) = 1
See if you can simplify your question using these equations, and arrive at an answer (which I think is not 0).
I am not sure how I am supposed to do it. does that mean the limit is 6?
Lim 3x^2 csc^2 can be written
x->0
Lim 3x^2/sin^2x
x->0
You can get the limit as x->0 using L'Hopital's rule: It is the ratio of the derivatives of numerator and denominator. You have to apply it twice here, since the first derivatives are also 0/0
Lim 3x^2/sin^2x
x->0
= Lim 6x/2sinxcosx
x->0
=Lim 6x/(sin 2x)
x->0
= 6/(2cos0) = 3
What Amrit suggested you do is to use the follwing rule:
Suppose Lim x ---> a of f(x) = y
Then Lim x ---> a of g(f(x)) = g(y)
if the function g(x) is continuous at the point x = y.
In this case you take f(x) to be
sin(x)/x and g(x) = 3 x^2. Because g(x) is a continuous function, the limit is 3*(1)^2 = 3.
Correction: f(x) = x/sin(x).
To find the limit as x approaches 0 of the function 3x^2csc^2(x), we can use basic limit properties and trigonometric identities.
First, let's simplify the expression using the trigonometric identity csc^2(x) = 1/sin^2(x):
3x^2csc^2(x) = 3x^2/(sin^2(x))
Now, let's focus on the denominator. As x approaches 0, sin(x) also approaches 0. We know that sin(x)/x approaches 1 as x approaches 0 (a well-known limit). Therefore, sin^2(x)/x^2 also approaches 1.
Now, let's substitute this result back into the original expression:
3x^2/(sin^2(x)) = 3x^2 * x^2/(sin^2(x) * x^2)
Simplifying further:
= 3x^4/(sin^2(x) * x^2)
Now, since sin^2(x)/x^2 approaches 1 and x^2/x^2 = 1, we can rewrite the expression as:
= 3x^4 * (1/1)
= 3x^4
Now, if we take the limit as x approaches 0, we get:
lim(x->0) 3x^4 = 0
Therefore, the answer is 0.