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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 -

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

  • Math -

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

  • Math -


    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 .

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