Posted by Neee on Saturday, August 14, 2010 at 9:38am.
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 dont 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 .
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