Calculus

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Evaluate each limit. If it exists.

a) Lim x^3 + 1
x->1 -------
x + 1

b) Lim 3x^2 - x^3
x->0 ----------
x^3 + 4x^2

c) Lim 16 - x
x->16 ---------
(√x) - 4

  • Calculus -

    do you mean x --> -1? Otherwise, there's no problem evaluating the fraction.
    x^3+1 = (x+1)(x^2-x+1)
    divide out the (x+1) to get
    limit = 3

    or, using L'Hopital's rule,

    lim (3x^2)/1 = 3



    x^2(3-x) / x^2(x+4) = (3-x)/(x+4) --> 3/4

    lim (16-x)/(√x-4) = lim (-1)/(1/(2√x)) = -1/(1/8) = -8

    or, divide out the √x-4, since 16-x = -(√x-4)(√x+4) --> -8

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