Recent content by moloko

  1. M

    Solving Trig Sub in Homework: Integral of (16x-2x^2-23)^1/2 dx

    Some work... integral signs before everything a=sqrt(4.5) x-4 = asecT dx=asecTtanTdTsqrt[-2*(asecT)^2-4.5] * sqrt[4.5] * secTtanTdT sqrt[-2*(4.5sec^2(T)-4.5)*4.5] * secTtanTdT sqrt[(-40.5)(tan^2(T)] * secTtanTdT Subbing T=arcsec((x-4)/a) sqrt[(-40.5)(tan^2(arcsec(T))] *...
  2. M

    Solving Trig Sub in Homework: Integral of (16x-2x^2-23)^1/2 dx

    Homework Statement integral of (16x-2x^2-23)^1/2 dx Homework Equations The Attempt at a Solution First off it seems clear that I must complete the square within the radical. This gives ((x-4)^2-4.5)^1/2 From there I seem to be lost, fumbling through my book for any kind of...
  3. M

    L'Hopital's Rule for Solving Limits

    I apologize, that was a careless mistake. I'll post my solution if it's of any help to anyone... the numerator becomes 2(sec^2(x)tan^2(x) + sec^2(x)), which of course goes to 2 when pushed to 0 leaving 2/-8, or -1/4 Thank you so much!
  4. M

    L'Hopital's Rule for Solving Limits

    You are correct and this does give me a non 0/0 answer but the answer is wrong according to the book. It should be -1/4. My next derivative is: (2*(2sec^2(x)tan(x))*tan(x))/(-8cos(2x)) In simpler terms (4sec^2(x)tan^2(x))/(-8cos(2x)) The numerator becomes 0 with the limit and the...
  5. M

    L'Hopital's Rule for Solving Limits

    Homework Statement lim as x->0 (tan(x)-x)/(sin2x-2x) Homework Equations L'Hopitals rule states that if the limit reaches 0/0, you can take the derivative of the top and the bottom until you get the real limit. The Attempt at a Solution (sec^2(x)-1)/(2cos2x-2) still 0/0...
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