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Math Help - Hard limit problem

  1. #1
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    Hard limit problem

    <br />
\displaystyle\lim_{n\to\infty}(\frac{n^2+3}{n^2+1}  )^{n^{2}+2}<br /> <br />

    Without the n^2+2 in exponent would be easy. But how do we get rid of that?

    Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

    The result should be e^2 or something like that.
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  2. #2
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    Quote Originally Posted by Nforce View Post
    <br />
\displaystyle\lim_{n\to\infty}(\frac{n^2+3}{n^2+1}  )^{n^{2}+2}<br /> <br />

    Without the n^2+2 in exponent would be easy. But how do we get rid of that?

    Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

    The result should be e^2 or something like that.
    \displaystyle\lim_{n\to\infty}\left[\left(\frac{n^2+3}{n^2+1}\right)^{n^2}\cdot\left(\  frac{n^2+3}{n^2+1}\right)^2\right]

    \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n  ^2+1}\right)^{n^2}\cdot\lim_{n\to\infty}\left(\fra  c{n^2+3}{n^2+1}\right)^2

    Does this help you?
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  3. #3
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    Quote Originally Posted by Nforce View Post
    <br />
\displaystyle\lim_{n\to\infty}(\frac{n^2+3}{n^2+1}  )^{n^{2}+2}<br /> <br />

    Without the n^2+2 in exponent would be easy. But how do we get rid of that?

    Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

    The result should be e^2 or something like that.
    \displaystyle\left(\frac{n^2+3}{n^2+1}\right)^{n^2  +2}=\left(\frac{(n^2+1)+2}{n^2+1}\right)^{(n^2+1)+  1

    =\displaystyle\left[\left(1+\frac{2}{n^2+1}\right)^{\frac{n^2+1}{2}}\r  ight]^2\left(1+\frac{2}{n^2+1}\right)


    Now you can use

    \displaystyle\ e=\lim_{x\ \to \infty}\left(1+\frac{1}{x}\right)^x
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  4. #4
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    Hello, Nforce!

    Are you allowed to use L'Hopital?


    \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n  ^2+1}\right)^{n^{2}+2}

    \text{We have: }\;y \;=\;\left(\dfrac{n^2+3}{n^2+1}\right)^{n^2+2}\;\;  \to\;\;1^{\infty}

    \text{Take logs: }\;\ln y \;=\;\ln\left(\dfrac{n^2+3}{n^2+1}\right)^{n^2+2} \;=\;(n^2+2)\ln\left(\dfrac{n^2+3}{n^2+1}\right) \;\;\to\;\;\infty \cdot 0

    . . . . . . . . . \ln y \;=\;\dfrac{\ln(n^2+3) - \ln(n^2+1)}{(n^2+2)^{-1}} \;\;\to\;\;\dfrac{\infty - \infty}{0}


    \text{Apply l'Hopital: }\;\ln y \;=\;\dfrac{\dfrac{2n}{n^2+3} - \dfrac{2n}{n^2+1}}{-\dfrac{2n}{(n+2)^2}} \;=\;\dfrac{2n\left(\dfrac{n^2+1 - n^2-3}{(n^2+3)(n^2+1)}\right)}{-\dfrac{2n}{(n^2+2)^2}}

    . . . . . . . . . . . . . . \ln y \;=\;\dfrac{\dfrac{2n(-2)}{(n^2+3)(n^2+1)}}{\dfrac{-2n}{(n^2+2)^2}} \;=\;\dfrac{2(n^2+2)^2}{(n^2+3)(n^2+1)}


    Divide numerator and denominator by n^4

    . . . . \ln y \;=\; \dfrac{2\left(1 + \frac{2}{n^2}\right)^2} {\left(1 + \frac{3}{n^2}\right)\left(1 + \frac{1}{n^2}\right)}


    Then: . \displaystyle \lim_{x\to\infty} \ln y \;=\;\lim_{x\to\infty} \dfrac{2\left(1 + \frac{2}{n^2}\right)^2} {\left(1 + \frac{3}{n^2}\right)\left(1 + \frac{1}{n^2}\right)} \;=\;\frac{2(1+0)^2}{(1+0)(1+0)} \;=\;2<br />



    \displaystyle \text{Since }\lim_{x\to\infty}\ln y \;=\;2,\,\text{ then: }\lim_{x\to\infty} y \;=\;e^2

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  5. #5
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    Yes we have used the L'hopital. But the thing is that we never solved limits with "n" in exponent. I feel sorry for you soroban, thanks for help but I don't understand the whole process, what does that arrow mean? and why did you write 1^infinity.


    I think this is something that I understand more. Just simple math rules.
    \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n  ^2+1}\right)^{n^2}\cdot\lim_{n\to\infty}\left(\fra  c{n^2+3}{n^2+1}\right)^2


    But I have some question tho.

    Could I rewrite that into this, and solve it

    \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n  ^2+1}\right)^{n^2}\cdot\left(\lim_{n\to\infty}\fra  c{n^2+3}{n^2+1}\right)^2

    But the first term has still n in the exponent. Oh man, can I write it into a fraction someway?
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  6. #6
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    You could simplify the process if you are unfamiliar with the limit expression for e^x

    y=n^2+2

    \displaystyle\left(\frac{n^2+3}{n^2+1}\right)^{n^2  +2}=\left(\frac{y+1}{y-1}\right)^y

    Take natural logs

    \displaystyle\ lnL=\lim_{y \to \infty}ln\left(\frac{y+1}{y-1}\right)^y=\lim_{y \to \infty}yln\left(\frac{y+1}{y-1}\right)=\lim_{y \to \infty}\frac{ln\left(\frac{y+1}{y-1}\right)}{\left(\frac{1}{y}\right)}

    which \rightarrow\frac{0}{0}

    Use L'Hopital's Rule, differentiate with the chain rule (and the quotient rule)....

    \displaystyle\lim_{y \to \infty}\frac{\frac{y-1}{y+1}\left[\frac{(y-1)(1)-(y+1)(1)}{(y-1)^2}\right]}{\left(-\frac{1}{y^2}\right)}

    and

    \displaystyle\frac{y-1}{y+1}\left[\frac{-2}{(y-1)^2}\right]\left(-y^2}\right)

    =\displaystyle\frac{(y-1)2y^2}{(y+1)(y-1)^2}=\frac{2y^2}{(y+1)(y-1)}=\frac{2y^2}{y^2-1}

    \displaystyle\lim_{y \to \infty}\frac{2y^2}{y^2-1}=2

    L=e^2
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  7. #7
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    I take y_n = n^2 + 1. The piece inside the exponent becomes

    <br />
\left(1 + \frac{2}{y_n}\right)<br />

    so the full expression is

    <br />
\left(1 + \frac{2}{y_n}\right)^{y_n} \cdot \left(1 + \frac{2}{y_n}\right)<br />

    Let y_n go to infinity and get e^2.

    This is the same as Archie's solution, but prettier to look at because we don't have a bunch of ugly things floating around
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  8. #8
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    Need Help

    Can anyone tel me?
    what is the value of tanh(x)/x when x tends to infinity and what is the value of sinh(x)/x when x tends to infinity?
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  9. #9
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    Quote Originally Posted by skullface View Post
    Can anyone tel me?
    what is the value of tanh(x)/x when x tends to infinity and what is the value of sinh(x)/x when x tends to infinity?
    start a new problem w/ a new thread ... don't tag, please.
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  10. #10
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    Respect!, Now I understand, thanks a lot.
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