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Math Help - limits

  1. #1
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    limits

    please help

    evaluate lim as x goes to infinity (1+(a/x))^x
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  2. #2
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    Quote Originally Posted by Gigabyte View Post
    please help

    evaluate lim as x goes to infinity (1+(a/x))^x

    Hint:


    \left(\lim_{x \to \infty} \left(1+\frac{a}{x}\right)^\frac{x}{a}\right)^a = e^a
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  3. #3
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    Hello,

    Quote Originally Posted by Gigabyte View Post
    please help

    evaluate lim as x goes to infinity (1+(a/x))^x
    \lim_{x \to \infty} {\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}=\lim_{x \to \infty} x \ln \left(1+\frac ax\right)

    Change the variable : t=\frac ax \implies x=\frac at

    When x \to \infty, t \to 0


    \lim_{x \to \infty} x \ln \left(1+\frac ax\right)=\lim_{t \to 0} \ln (1+t) \cdot \frac at=\lim_{t \to 0} a \cdot \frac{\ln(1+t)}{t}=a \cdot \lim_{t \to 0} \frac{\ln(1+t)-\ln(1)}{t}

    But \lim_{t \to 0} \frac{\ln(1+t)-\ln(1)}{t} is, according to the definition of the derivative number, \left(\ln(1+t)\right)'_{t=0}=\frac{1}{1+t} \bigg|_{t=0}=1


    Therefore, the limit is \lim_{x \to \infty} {\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}=a \cdot 1=\boxed{\color{red}a}

    But \left(1+\frac ax\right)^x=e^{\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}


    ---> \boxed{\lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a}
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  4. #4
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    thanks heaps for that bru.

    Is 1 line wrong though.
    lim as t tends to 0

    Isn't it supposed to be:
    ln(1+t)^(a/t)

    Not:
    ln(1+t)*(a/t) as you wrote it??
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  5. #5
    Moo
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    Quote Originally Posted by Gigabyte View Post
    thanks heaps for that bru.

    Is 1 line wrong though.
    lim as t tends to 0

    Isn't it supposed to be:
    ln(1+t)^(a/t)

    Not:
    ln(1+t)*(a/t) as you wrote it??
    \ln (a^b)=b \ln(a)

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  6. #6
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    yip that makes sense. Sorry one of those Rookie errors.

    Why do you get?
    (ln(1+t)-ln(1))/t
    When you take (a) out the front of the limit. You take (a) out the front cos it's a constant right?

    Then after that i kinda didn't get the next part.
    "But is, according to the definition of the derivative number, "
    Could you please explain that because i think i'm getting.
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  7. #7
    Moo
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    Quote Originally Posted by Gigabyte View Post
    yip that makes sense. Sorry one of those Rookie errors.

    Why do you get?
    (ln(1+t)-ln(1))/t
    \ln(1)=0, so I could add it, right ?

    When you take (a) out the front of the limit. You take (a) out the front cos it's a constant right?
    Yes

    Then after that i kinda didn't get the next part.
    "But is, according to the definition of the derivative number, "
    Could you please explain that because i think i'm getting.
    Have look here : Derivative - Wikipedia, the free encyclopedia
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by Gigabyte View Post
    Then after that i kinda didn't get the next part.
    "But is, according to the definition of the derivative number, "
    Could you please explain that because i think i'm getting.

    Are you introduced to differentiation? If not then you have to use limit definition of \lim_{n \to \infty} \left(1+\frac1{n}\right)^n = e

    So follow what I said in post 2.
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  9. #9
    Moo
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    Quote Originally Posted by Isomorphism View Post
    Are you introduced to differentiation? If not then you have to use limit definition of \lim_{n \to \infty} \left(1+\frac1{n}\right)^n = e

    So follow what I said in post 2.
    Is it possible that one is taught this limit before doing differentiation ?

    LOL ! I forgot... by the way, his/her previous thread was about implicit differentiation
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  10. #10
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    Quote Originally Posted by Moo View Post
    Is it possible that one is taught this limit before doing differentiation ?
    Ya obviously, We are taught limits first... how else will you learn the definition of differentiation?


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    Quote:
    Originally Posted by Isomorphism View Post
    Are you introduced to differentiation? If not then you have to use limit definition of \lim_{n \to \infty} \left(1+\frac1{n}\right)^n = e

    So follow what I said in post 2.
    Is it possible that one is taught this limit before doing differentiation ?

    LOL ! I forgot... by the way, his/her previous thread was about implicit differentiation
    Oh ok then
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  11. #11
    Moo
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    Quote Originally Posted by Isomorphism View Post
    Ya obviously, We are taught limits first... how else will you learn the definition of differentiation?
    I was talking about this limit, not limits in general...
    I don't see how you can prove this limit with tools that are taught before differentiation...
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  12. #12
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    Yea i have been introduced to differentiation. However i don't really understand that line. I tried to read about it but got more confused.

    So from post 8 would you get????
    lim as x tents to infinity
    ln[(1+a/x)^(x/a)]^a = e^a
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  13. #13
    Moo
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    Quote Originally Posted by Gigabyte View Post
    Yea i have been introduced to differentiation. However i don't really understand that line. I tried to read about it but got more confused.
    You just have to know that :

    \lim_{t \to 0} \frac{f(a+t)-f(a)}{t}=f'(a)

    Here, a=1 and f(x)=\ln(1+t) \implies f'(x)=\frac{1}{1+t}

    So from post 8 would you get????
    lim as x tents to infinity
    ln[(1+a/x)^(x/a)]^a = e^a
    yeah, this is the result
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  14. #14
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    wow that was a mammoth step. But i think i have caught up to you. Now i'm stuck again lol.

    What is the derivative number, and how does it help to solve the next line of working in post 3?? Otherwise i think i've got it.

    Thank you so much for your help so far. It's been awesome, i'm really learning alot.
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  15. #15
    Moo
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    The derivative number in x=a is actually the value of the derivative at this point x=a.
    It's the same as taking the value of the derivative function in a.

    This is the very first definition of derivative that you should have learnt


    Is it all clear now ?
    Last edited by Moo; May 22nd 2008 at 07:07 AM.
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