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Math Help - lim (1+1/x)^x as x->infinity

  1. #1
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    lim (1+1/x)^x as x->infinity

    Good afternoon,

    I am working on a problem that asks to find the following limit:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    This should be equal to e, but I am having some difficulty remembering why.

    I checked WolframAlpha and was provided with the following:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    Indeterminate form of type 1^\infty. Transform using
     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />
=  <br />
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})<br />


    Where did this come from?
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  2. #2
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    Using Napierís inequality it is easy to show that:
    \displaystyle \frac{1}{n+1}\le\log \left( {1 + \frac{1}{n}} \right) \leqslant \frac{1}{n} .

    So e^{\frac{1}{{n + 1}}}  \leqslant \left( {1 + \frac{1}{n}} \right) \leqslant e^{\frac{1}{n}} \; \Rightarrow \;e^{\frac{n}{{n + 1}}}  \leqslant \left( {1 + \frac{1}{n}} \right)^n  \leqslant e

    From that the limit is clear.
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  3. #3
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    Quote Originally Posted by MechEng View Post
    Good afternoon,

    I am working on a problem that asks to find the following limit:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    This should be equal to e, but I am having some difficulty remembering why.

    I checked WolframAlpha and was provided with the following:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    Indeterminate form of type 1^\infty. Transform using
     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />
=  <br />
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})<br />


    Where did this come from?
    If you let y=\left(1+\frac{1}{x}\right)^x Then take the log of both sides you can write:

    ln(y)=x\ln\left(1+\frac{1}{x}\right)

    Then take the limit of both sides.

    ln(\lim_{x->\infty}y)=\lim_{x->\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac  {1}{x}}

    Now use l-hospitals rule:

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  4. #4
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    See this thread

    CB
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  5. #5
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    Quote Originally Posted by MechEng View Post
    Good afternoon,

    I am working on a problem that asks to find the following limit:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    This should be equal to e, but I am having some difficulty remembering why.

    I checked WolframAlpha and was provided with the following:

     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />

    Indeterminate form of type 1^\infty. Transform using
     <br />
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x<br />
=  <br />
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})<br />


    Where did this come from?
    The Binomial Expansion is

    (a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+....+\binom{n}{n}a^{n-n}b^n

    The Taylor Series expansion for e^x is

    e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2  !}+\frac{x^3}{3!}+.....

    Using the Binomial Expansion for \left(1+\frac{1}{x}\right)^x

    \left(1+\frac{1}{x}\right)^x=\binom{x}{0}+\binom{x  }{1}\frac{1}{x}+\binom{x}{2}\frac{1}{x^2}+\binom{x  }{3}\frac{1}{x^3}+......

    =1+1+\frac{x(x-1)}{2!}\frac{1}{x^2}+\frac{x(x-1)(x-2)}{3!}\frac{1}{x^3}+....

    =1+1+\frac{x^2-x}{2!x^2}+\frac{x^3-3x^2+2x}{3!x^3}+.....

    Using the fact that \displaystyle \lim_{x\ \rightarrow\ \infty}\frac{1}{x^n}=0,\ n=1,\ 2,\ 3,\ 4,\ ......

    and all the powers of x will be in the denominator position above,
    therefore as x goes to infinity, it reduces to

    \displaystyle \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right  )^x=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\fr  ac{1}{5!}+.....=e^1
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  6. #6
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    I am still somewhat confused by this. I can see that e is defined as e=lim_{n\rightarrow\infty}(1+\frac{1}{n})^n.

    I think I might be having another mental lapse with respect finding this limit. I am working on sequences, so is there some corelation there that I am missing?
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  7. #7
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    Quote Originally Posted by adkinsjr View Post
    If you let y=\left(1+\frac{1}{x}\right)^x Then take the log of both sides you can write:

    ln(y)=x\ln\left(1+\frac{1}{x}\right)

    Then take the limit of both sides.

    ln(\lim_{x->\infty}y)=\lim_{x->\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac  {1}{x}}

    Now use l-hospitals rule:



    Ah...

    I think I follow this, however...

    When we take the log of both sides and get:

    ln(y)=x\ln\left(1+\frac{1}{x}\right)

    Why are our new limits not:

    ln(\lim_{x->\infty}y)=xln\lim_{x->\infty}(1+\frac{1}{x}\right)

    Is the ln treated as a constant?

    I hope my question is not too rediculous.

    PS... I would love to watch your youtube video, but youtube is blocked at my place of work.
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  8. #8
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    Quote Originally Posted by MechEng View Post
    Ah...

    I think I follow this, however...

    When we take the log of both sides and get:

    ln(y)=x\ln\left(1+\frac{1}{x}\right)

    Why are our new limits not:

    ln(\lim_{x->\infty}y)=xln\lim_{x->\infty}(1+\frac{1}{x}\right)

    Is the ln treated as a constant?

    I hope my question is not too rediculous.

    PS... I would love to watch your youtube video, but youtube is blocked at my place of work.
    Firstly,

    y=\left(1+\frac{1}{x}\right)^x

    If we take the natural logarithms of both sides, we will be able to handle the "index" x before evaluating the limit using L'Hopital's rule.

    log_ey=log_e\left(1+\frac{1}{x}\right)^x=xlog_e\le  ft(1+\frac{1}{x}\right)=\frac{1}{\left(\frac{1}{x}  \right)}log_e\left(1+\frac{1}{x}\right)=\frac{log_  e\left(1+\frac{1}{x}\right)}{\left(\frac{1}{x}\rig  ht)}

    This brings y into a form that can be evaluated as x\rightarrow\ \infty using L"Hopital's Rule, since if x\ \rightarrow\ \infty

    \frac{log_e\left(1+\frac{1}{x}\right)}{\left(\frac  {1}{x}\right)} evaluates to

    \frac{log_e1}{0}\ \rightarrow\ \frac{0}{0}

    Hence, differentiating numerator and denominator to apply L'Hopital's Rule..

    \frac{\frac{d}{dx}log_e\left(1+\frac{1}{x}\right)}  {\frac{d}{dx}\left(\frac{1}{x}\right)}=\frac{\frac  {1}{1+\frac{1}{x}}\left(-\frac{1}{x^2}\right)}{\left(-\frac{1}{x^2}\right)}

    =\frac{1}{1+\frac{1}{x}}

    Now this can be evaluated as x approaches infinity....

    log_ey=\frac{1}{1+0} as x\ \rightarrow\ \infty

    y=e^1

    Therefore..

    \displaystyle\ \lim_{x\rightarrow\ \infty}\left(1+\frac{1}{x}\right)^x=e
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  9. #9
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    First off, thank you all for your continued efforts here.

    Secondly, I apologize for being so slow to catch on here, requiring everyone to reiterate.

    However, I think I have a more clear and concise understanding of where I'm losing you...

    log_ey=log_e\left(1+\frac{1}{x}\right)^x


    =xlog_e\left(1+\frac{1}{x}\right)


    Right about here... how (or why, for that matter) did we end up with x=\frac{1}{(\frac{1}{x})} ?

    Are we simply jostling the terms about to allow us to apply L'Hopital's rule?


    =\frac{1}{\left(\frac{1}{x}\right)}log_e\left(1+\f  rac{1}{x}\right)


    =\frac{log_e\left(1+\frac{1}{x}\right)}{\left(\fra  c{1}{x}\right)}

    I appologize if this process has been frustrating... it has been very taxing for me.
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  10. #10
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    Quote Originally Posted by MechEng View Post
    First off, thank you all for your continued efforts here.

    Secondly, I apologize for being so slow to catch on here, requiring everyone to reiterate.

    However, I think I have a more clear and concise understanding of where I'm losing you...

    log_ey=log_e\left(1+\frac{1}{x}\right)^x


    =xlog_e\left(1+\frac{1}{x}\right)


    Right about here... how (or why, for that matter) did we end up with x=\frac{1}{(\frac{1}{x})} ?

    Are we simply jostling the terms about to allow us to apply L'Hopital's rule?


    =\frac{1}{\left(\frac{1}{x}\right)}log_e\left(1+\f  rac{1}{x}\right)


    =\frac{log_e\left(1+\frac{1}{x}\right)}{\left(\fra  c{1}{x}\right)}

    I appologize if this process has been frustrating... it has been very taxing for me.
    Yes, you could say we are jostling terms here in order to reach a suitable form to apply the rule.

    1=\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{x  }\right)}

    hence, x=\frac{x\left(\frac{1}{x}\right)}{\left(\frac{1}{  x}\right)}=\frac{\left(\frac{x}{x}\right)}{\left(\  frac{1}{x}\right)}

    =\frac{1}{\left(\frac{1}{x}\right)}

    Take your time to understand.
    There are numerous alternative ways to tackle the problem,
    including that very interesting inequality pointed out by Plato.
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