posted by Neee .
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)
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 .