Simplify:

f(x)=sqrt (16-x^4)/ sinx

This is to evaluate a limit, but I don't know if I can simplify the above equation. Because I tried to get rid of the radical in the numerator but I ended up with a radical in the denominator. >< Help

Post the entire limit question. Is the Sin in the sqrt or is it (sqrt(_))/sinx

Where is f(x)=(sqrt(16-4^4))/sinx continuous? Evaluate lim x-> pi/2 f(x)

sqrt(16-x^4)/sinx

as x>PI/2
Then limit is easy...
Lim = sqrt(16-(pi/2)^4)/1

I don't see the problem.

Now, where is it continous?

when 16-x^4 goes negative, there is a problem with a sqrt of a negative number, so when x=2, there is a disconituity as the f(x)=0 .

To simplify the given function f(x) = sqrt(16 - x^4) / sin(x), we can start by simplifying the numerator and the denominator separately.

For the numerator, sqrt(16 - x^4), we can recognize that 16 - x^4 can be factored using the difference of squares formula: a^2 - b^2 = (a + b)(a - b). Applying this formula in our case, we have:

16 - x^4 = (4 + x^2)(4 - x^2)

Next, we simplify the denominator, sin(x).

Now, the original function can be rewritten as:

f(x) = sqrt((4 + x^2)(4 - x^2)) / sin(x)

For the further simplification, we can apply two trigonometric identities:

1. sin^2(x) + cos^2(x) = 1
2. sin(2x) = 2sin(x)cos(x)

First, let's simplify the denominator further using the identity sin^2(x) + cos^2(x) = 1:

f(x) = sqrt((4 + x^2)(4 - x^2)) / sin(x)
= sqrt(16 - x^4) / sin(x)
= (sqrt(16 - x^4)) / (sqrt(sin^2(x) + cos^2(x)))

Next, we apply the identity sin(2x) = 2sin(x)cos(x) to the denominator:

f(x) = (sqrt(16 - x^4)) / (sqrt(sin^2(x) + cos^2(x)))
= sqrt(16 - x^4) / (sqrt(sin^2(x) + cos^2(x)))
= sqrt(16 - x^4) / (sqrt((2sin(x)cos(x))^2 + cos^2(x)))
= sqrt(16 - x^4) / (sqrt(4sin^2(x)cos^2(x) + cos^2(x)))

Finally, we can factor out cos^2(x) from the denominator:

f(x) = sqrt(16 - x^4) / (sqrt(4sin^2(x)cos^2(x) + cos^2(x)))
= sqrt(16 - x^4) / (sqrt(cos^2(x)(4sin^2(x) + 1)))

At this point, the function f(x) = sqrt(16 - x^4) / sin(x) cannot be simplified further, as we have factored out the common factors.