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Calculus I Questions

  1. Sep 28, 2005 #1
    1.) lim x --> 1(left) (x^2 + |x| -2)/(|1-x|)

    Since the limit is from the left, i made all absolute values pnegative, therefore numerator = x^2 - x - 2, and similarly denom. = (x-1). Then, by inspection, the limit would equal to + infinite.

    2.) Find numbers a and c such that lim x --> 0 x/(sqrt(ax + c) -3) = 2

    Multiplying first by the conjugate, and then assuming a value for either a or c (in my case, i let c = 9). Then, solving for a, i got 3. Is this assumption allowed/necessary?

    3.) lim (x-->0) f(x^2)/x^2, where f(0)=0 and f'(0)=3.

    No idea.

    4.) Evaluate f'(P), where f(P) = tan(3P + sinP))

    No idea.

    5.) Prove: If lim x--> 6 f(x)g(x) exists, then limit must be f(6)g(6).

    6.) Prove: If f is continuous at 5 and f(5)=2 and f(4)=3, then lim x-->2f(4x^2 - 11) =2


    Thank you for absolutely any input. These are only a selected few (out of very many) that i was unsuccessful at solving.

    Thank you again.
     
  2. jcsd
  3. Sep 28, 2005 #2
    #1 :
    1. the numerator is x^2 + x -2 because when x approaches 1 from the left, it is a positive number; and can factorized.
    2. simplify and i get -3.
     
  4. Sep 28, 2005 #3
    4) Have you learned derivatives yet?

    some things that may help you solve the problem

    a) tan(x) = sin(x)/cos(x)
    b) The two most important 'differentiation rules' here are the quotient and chain rule.
    c) d/dx(sinx) = cosx, d/dx(cosx) = -sinx where d/dx is the derivative with respect to x if you haven't seen that notation before

    Try posting anything you come up with
     
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