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Math Help - Limits: h approaching zero

  1. #1
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    Limits: h approaching zero

    Can someone please explain how to solve this?
    Thanks in advance.


    Let f(x)=x^3

    What number does [f(5+h)-f(5)]/h approach as h approaches zero?
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  2. #2
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    We have
    \lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}

    =\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times  5h^2-5^3}{h}

    =\lim_{h\rightarrow0}{3\times5^2+3\times5h}

    =3\times5^2=75
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  3. #3
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    Quote Originally Posted by thelostchild View Post
    We have
    \lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}

    =\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times  5h^2-5^3}{h}

    =\lim_{h\rightarrow0}{3\times5^2+3\times5h}

    =3\times5^2=75
    Would you mind explaining how you got: =\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times  5h^2-5^3}{h} ?
    I don't quite follow. And where did those 3's come from?
    Thank you.
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  4. #4
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    Quote Originally Posted by overduex View Post
    Would you mind explaining how you got: =\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times  5h^2-5^3}{h} ?
    I don't quite follow. And where did those 3's come from?
    Thank you.
    (5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    (5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.
    I really should be less asleap while doing it!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by overduex View Post
    ...I don't quite follow. And where did those 3's come from?
    Thank you.
    just to add a bit to what Mr F said,

    the hard way: write as (5 + h)(5 + h)(5 + h)

    and expand the brackets two at a time. that is, leave the last one and expand the first two. then expand the remaining two brackets


    the easy way: note that (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

    now plug in your values for a and b
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    (5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.
    Ok, so I expanded it but I still can't get the answer. Can you check for a mistake somewhere?
    =\lim_{h\rightarrow0}\frac{5+h\times 25+10h+h^2- 125}{h}
    =\lim_{h\rightarrow0}\frac{125+ 50h+5h+25h+ 10h+ 10h^2 + h^3 - 125}{h}
    =\lim_{h\rightarrow0}\frac{80h +10h^2 +h^3}{h}
    so all I factored out the h and ended up with

    ={80 + 10h + h^2}
    which would mean the limit would be approaching 80, but thats not the answer... Help? Please?
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  8. #8
    Junior Member Ziaris's Avatar
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    You expanded incorrectly. Use Jhevon's easy way by the binomial theorem to check your expansion. You're supposed to get

    \lim_{h\to 0}\frac{5^3 + 75h+15h^2+h^3 - 5^3}{h}.
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