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Math Help - Limits

  1. #1
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    Limits

    Could someone explain to me how to find the limits for the following two expressions as n approaches infinity:

    a) <br />
\frac{{5^{n + 2}  - 7^{n + 2} }}{{5^n  - 7^n }}<br />
    b) <br />
\frac{{n + ( - 2)^n }}{{n - ( - 2)^n }}<br />

    help is hugely appreciated.
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\mathop {\lim }\limits_{n \to \infty } \frac{{5^{n + 2}  - 7^{n + 2} }}<br />
{{5^n  - 7^n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{25\left( {\frac{5}<br />
{7}} \right)^n  - 49}}<br />
{{\left( {\frac{5}<br />
{7}} \right)^n  - 1}} = 49

    <br />
\begin{gathered}<br />
  \frac{{n - 2^n }}<br />
{{n + 2^n }} < \frac{{n + \left( { - 2} \right)^n }}<br />
{{n - \left( { - 2} \right)^n }} < \frac{{n + 2^n }}<br />
{{n - 2^n }} \hfill \\<br />
  \mathop {\lim }\limits_{n \to \infty } \frac{{n - 2^n }}<br />
{{n + 2^n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 2^n }}<br />
{{n - 2^n }} =  - 1 \hfill \\ <br />
\end{gathered}

    thus by the squeeze theorem:

    <br />
\mathop {\lim }\limits_{n \to \infty } \frac{{n + \left( { - 2} \right)^n }}<br />
{{n - \left( { - 2} \right)^n }} =  - 1<br />
    Last edited by Peritus; November 1st 2008 at 06:49 AM.
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  3. #3
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    Squeeze theorem? I've not heard of that, but I was under the impression that I was suppose to solve the question using l'Hopital's rule, except I kept getting the limit of the derivatives to be 0 or infinity and the whole limit is still indeterminate. Does that mean I can't use that rule here?
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by free_to_fly View Post
    I was under the impression that I was suppose to solve the question using l'Hopital's rule
    Stop thinking mechanically, first look at the problem at hand and think.

    As you can see this is a limit of a sequence - a discrete set of values. The derivative is a continues variable operator, whereas in this case we're given a sequence - which is a discrete set of values, so the derivative has absolutely no meaning when applied to the series. This is exactly what happens when people try to apply theorems, operators, transformations... without understanding their meaning and the necessary conditions for their application. I can't blame you because I was too one of those people, but I've learned the hard way that I had to change.
    We could apply the discrete version of L'Hospital rule to the sequence (The derivative is replaced with a difference), but unfortunately the necessary conditions do not hold.

    Squeeze theorem - Wikipedia, the free encyclopedia
    Last edited by Peritus; November 1st 2008 at 07:51 AM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Peritus View Post
    <br /> <br />
{{n + 2^n }} < \frac{{n + \left( { - 2} \right)^n }}<br />
{{n - \left( { - 2} \right)^n }} < \frac{{n + 2^n }}<br />
{{n - 2^n }} \hfill \\<br />
\mathop {\lim }\limits_{n \to \infty } \frac{{n - 2^n }}<br />
{{n + 2^n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 2^n }}<br />
{{n - 2^n }} = - 1 \hfill \\ <br />
\end{gathered}

    thus by the squeeze theorem:

    <br />
\mathop {\lim }\limits_{n \to \infty } \frac{{n + \left( { - 2} \right)^n }}<br />
{{n - \left( { - 2} \right)^n }} = - 1<br />
    Alternatively you could rewrite the limit as

    \lim_{n\to\infty}\frac{\frac{n}{(-2)^n}+1}{\frac{n}{(-2)^n}-1}

    And it is very well known that the exponential function grows much faster than any polynomial expression in other words \left(a>1\right)^n\succ{x^n}

    Therefore the limit is

    \frac{0+1}{0-1}=-1
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  6. #6
    Senior Member Peritus's Avatar
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    Oh is that so?

    Then would you care to tell me what is the limit of:

    <br />
\mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^n

    Incidentally your false arguments gives the correct result in this case, but it is nonetheless a false one.

    One could use your argument had the sequence was:

    <br />
\frac{{n + 2^n }}<br />
{{n - 2^n }}<br />
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  7. #7
    Moo
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    Quote Originally Posted by Peritus View Post
    Oh is that so?

    Then would you care to tell me what is the limit of:

    <br />
\mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^n

    Incidentally your false arguments gives the correct result in this case, but it is nonetheless a false one.

    One could use your argument had the sequence was:

    <br />
\frac{{n + 2^n }}<br />
{{n - 2^n }}<br />
    Hmm well, for \frac{(-2)^n}{n}, mathstud is correct :

    \forall ~ n \geq 0, ~ -1 \leq (-1)^n \leq 1

    Therefore \frac{-1}{n} \leq \frac{(-1)^n}{n} \leq \frac 1n

    so \lim_{n \to \infty} \frac{-1}{n} \leq \lim_{n \to \infty} \frac{(-1)^n}{n} \leq \lim_{n \to \infty} \frac 1n

    \implies \lim_{n \to \infty} \frac{(-1)^n}{n}=0
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  8. #8
    Senior Member Peritus's Avatar
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    You're absolutely right, but mathstud's reasoning was entirely different and false.

    And it is very well known that the exponential function grows much faster than any polynomial expression in other words
    Anyway maybe I misunderstood him, in that case except my apology.
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  9. #9
    Moo
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    Quote Originally Posted by Peritus View Post
    You're absolutely right, but mathstud's reasoning was entirely different and false.



    Anyway maybe I misunderstood him, in that case except my apology.
    Yeah, thinking back about it, I'm just finding what he said nonsense.
    Because there is no link when he "concludes" by :
    Therefore the limit is
    \left(a>1\right)^n\succ{x^n}
    In fact, if |a|>1, then \lim_{n \to \infty} \frac{|a|^n}{n}=+ \infty and \lim_{n \to \infty} \frac{a^n}{n} is undefined if a<0.

    If |a|\le 1, \lim_{n \to \infty} \frac{a^n}{n}=0


    I think this pretty sums up the stuff, without using weird notations like \succ
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Peritus View Post
    Oh is that so?

    Then would you care to tell me what is the limit of:

    <br />
\mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^n

    Incidentally your false arguments gives the correct result in this case, but it is nonetheless a false one.

    One could use your argument had the sequence was:

    <br />
\frac{{n + 2^n }}<br />
{{n - 2^n }}<br />
    Sorry everyone! Once again my famous carelessness has caused a problem. Let me be more explicit.

    This is no proof but it gets the idea off. Consider

    \lim_{x\to\infty}\frac{x^a}{a^x}\quad|a|>1

    It can be seen that this is an infinity over infinity case and we can apply L'hospitals. In fact it will be infinity over infinity until the a+1th derivative. In other words

    \lim_{x\to\infty}\frac{x^a}{a^x}=\underbrace{\cdot  s}_{a\text{ times}}=\lim_{x\to\infty}\frac{a!}{\ln(a)^aa^x}=0

    Therefore \lim_{x\to\infty}\frac{x^a}{a^x}=0\quad\forall{|a|  >1}

    And if your grievance was that I had a>1 and not |a|>1 just make (-a)^n=(-1)^na^n and use the Squeeze theorem to deduce the limit is still zero.
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  11. #11
    Junior Member toraj58's Avatar
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    Talking

    = \lim_{n\to\infty} \frac {7^{n+2}(\overbrace{(\frac {5^{n+2}}{7^{n+2}})}^0 - 1)}{7^n(\underbrace{(\frac {5^n}{7^n})}_0 - 1)}

    = \frac {7^{n+2}}{7^n} = \frac {\rlap{\color{red}/}{7^n}\times 7^2}{\rlap{\color{red}/}{7^n}} = 49
    Last edited by toraj58; November 2nd 2008 at 06:59 AM.
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