Whats the limit of?

Lim square root of ( (x^2+ax) - square root of(x^2+bx) )

x increase without bound (+ infinity)

3 Answers

  • Mark P
    Lv 5
    1 decade ago
    Favorite Answer

    The limit is infinity.


    As x approaches infinity, you can throw out the ax and bx terms because x^2 dominates.

    So sqrt(x^2-sqrt(x^2)) = sqrt(x^2 - x)

    Again, the x^2 dominates, so you simply have:

    lim(x->infinity) of x

    which is simply infinity.

    Now, if you misplaced your parentheses in the problem statement, the answer is more interesting.


    lim(x->inf) of sqrt(x^2+ax) is x+a/2, since (x+a/2)^2 = (x^2+ax+1)

    lim(x->inf) of sqrt(x^2+bx) is x+b/2

    So lim(x->inf) of the difference is (x+a/2)-(x+b/2) or (a-b)/2.

    This is confirmed with calculations using very large x and small a and b.

  • 1 decade ago

    I believe it is (a-b)/2.

    My reasoning:

    x^2 + ax + (a/2)^2 = (x+a/2)^2 and

    x^2 + bx + (b/2)^2 = (x+b/2)^2

    So the limit looks like limit:

    sqrt[(x+a/2)^2] - sqrt[(x+b/2)^2] =

    x+a/2 - (x+b/2) = a/2 - b/2

    It's not formal, but if I had to put money on this problem, I would put it on (a-b)/2

  • Mafia
    Lv 4
    1 decade ago

    if lim of equation when x approaches infinity, then limit is rad a - rad b.

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