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calculus (limits)

lim h>0 sqrt(1+h)-1/h

not sure how to factor this; not allowed to use L'Hopital's Rule. (that isn't taught at my school until Calc II & I'm in Calc I).

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  1. I bet you mean

    lim h>0 [sqrt(1+h)-1] /h

    One way:
    I guess you know how to take derivatives
    d/dx (x)^.5 = .5 (x^-.5)
    (of course right there you can say
    x^-.5 at x = 1 so your result is
    .5/1 = .5
    so
    d (x^.5) = .5 x^-.5 dx
    so
    d (sqrt (x)) dx = .5 dx/sqrt(x)

    d sqrt(1+h) dx = .5 dx/sqrt(1+h)
    if h -->0
    d sqrt(1+h)dx = .5 sqrt(1)h/sqrt(1)
    = .5 h
    so sqrt(1+h) --> 1 + .5 h
    sqrt(1+h)-1 > .5 h
    divide by h and get
    .5

    series for sqrt(x) for x approximately 1 is

    =

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  2. I do know how to take derivatives however the course I'm in hasn't taught them yet (long story) so I can't use derivatives in finding the solution; I have to factor.

    So perhaps a better question is: how do I factor the above expression in a way that will allow me to find the limit?

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  3. Taylor series for f(x + a ) for x approximately 1 is

    f(x)= f(1) + f'(1)(x-a) + f''(x-a)^2/2 ...
    for f(x) = sqrt (x)
    f(1) = (1)^.5 = 1
    f'(1) = .5 (1)^-.5 = .5
    f"(x) = -.25(1)^-1.5 etc = -.25

    f(x+h) = 1 + .5*h -.25 h^2 /2 etc
    subtravt 1
    .5h -.125 h^2 tc
    divide by h and let h-->0
    .5 - .125 h --> .5

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  4. factor (x+h)^.5 ?

    I do not know how
    Somehow you have to know that sqrt(1+h) ---> 1 + .5 h +.....

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  5. Perhaps someone else knows how

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  6. rationalize the numerator, that is, multiply top and bottom by (√(1+h) + 1)/(√(1+h) + 1)

    lim (√(1+h) - 1)/h as h --->0
    = lim (√(1+h) - 1)/h (√(1+h) + 1)/(√(1+h) + 1)
    = lim (1+h -1)/[h( (√(1+h) + 1)/h)]
    = lim 1/( (√(1+h) + 1)
    = 1/(1+1) = 1/2

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