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April 1, 2015

April 1, 2015

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

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

- Math -
**bobpursley**, Saturday, August 14, 2010 at 9:45amPost the entire limit question. Is the Sin in the sqrt or is it (sqrt(_))/sinx

- Math -
**Neee**, Saturday, August 14, 2010 at 9:52amWhere is f(x)=(sqrt(16-4^4))/sinx continuous? Evaluate lim x-> pi/2 f(x)

- Math -
**bobpursley**, Saturday, August 14, 2010 at 11:35amsqrt(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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