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ppdj
ppdj

lim x->infinity ((sqrt(x^2+5))-(sqrt(x^2+3)))

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Posted by ppdj on Thursday October 2, 2008 at 10:53 AM and tagged with calculus, math.

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  1. james0tucson

    The answer is Zero, but I suspect you knew that before you posed the question.  

    We can simplify this problem without resorting to calculus.  Recall that

    lim x->infty(f(x) + g(x)) = lim x->infty(f(x)) + lim x->infty(g(x))

    Now it is easy to see that sqrt(x^2+5) approaches infinity with x, and so does sqrt(x^2+3).  Since the limits are equal, the difference of the limits is zero.

    Sources:
    http://web.mit.edu/wwmath/calculus/summary.html

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    Posted by james0tucson on Tuesday October 21, 2008 at 5:48 PM

  2. neela
    neela Teacher
    Graduate School

    An alternate proof using sandwich concept:

    To find Limitx->infinity {sqrt(x^2+5)-sqrt(x^2+3)}

    Let us look at what is sqrt(x^2+5)-sqrt(x^2+3). Let it be called f(x). Let us rationalise the numerator , by writing in numerator and denominator form.

    We see that f(x) > 0 for all x , as the square root of greater positive number is greater than that of a smaller positive number.

    f(x)={sqrt(x^2+5)-{sqrt(x^2+3) {sqrt(x^2+5)-{sqrt(x^2+3)}/{sqrt(x^2+5)+sqrt(x^2+3)}

    ={(x^2+5)-(x^2+3)}/{sqrt(x^2+5)+sqrt(x^2+3)}

    =2/{sqr(x^2+5)+sqrt(x^2+3)}  <  2/(2x) < 1/x.

    Now taking the limits,

    Lim x-> infinity f(x) < lim x->infinity (1/x )=0

    but f(x)>0 for all x.

    Thus 0<f(x)<0. f(x) is sandwiching between zero below and a zero aprroaching value from above as x approaches infininty. Therefore, Lim x->infinity f(x) = 0

    Therefore, Limit x->infinity {sqrt(x^2+5)-sqrt(x^2+3)} = 0.

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    Posted by neela on Thursday May 28, 2009 at 12:04 PM