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Math Help - Indeterminate Forms

  1. #1
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    Indeterminate Forms

    Can someone tell how to answer this problem using indeterminate form?



    Thanks in advance..
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  2. #2
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    Quote Originally Posted by cutiemike1 View Post
    Can someone tell how to answer this problem using indeterminate form?



    Thanks in advance..
    Logs can help in such cases...

    L=\displaystyle\lim_{x \to 0}\left(1+x^2\right)^{\frac{4}{x}}

    \displaystyle\ log_eL=\lim_{x \to 0}log_e\left(1+x^2\right)^{\frac{4}{x}}=\lim_{x \to 0}\left[\frac{log_e\left(1+x^2\right)^4}{x}\right]

    and using L'Hopital's rule, since we have a 0/0 indeterminate form

    \displaystyle\ =\lim_{x \to 0}\frac{4\left(1+x^2\right)^32x}{\left(1+x^2\right  )^4}

    \displaystyle\ =\lim_{x \to 0}\frac{8x\left(1+x^2\right)^3}{\left(1+x^2\right)  ^3\left(1+x^2\right)}

    \displaystyle\lim_{x \to 0}\frac{8x}{1+x^2}=0

    log_eL=0\Rightarrow\ e^0=L=1
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  3. #3
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    There is no reason not to take that "4" out of the exponent along with the "1/x":

    ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}
    and, by L'Hopital's rule
    ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}
    which is 0.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    There is no reason not to take that "4" out of the exponent along with the "1/x":

    ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}
    and, by L'Hopital's rule
    ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}
    which is 0.
    Thanks HallsofIvy.
    That's a huge improvement!
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Logs can help in such cases...
    I thought of that one but on my part I used natural logarithm. When I was solving it using natural logarithm I suddenly didn't know what must be done. This one will help me a lot in my final exam. MANY THANKS TO YOU!!
    Now my questions are:

    1. What might be the other cases which uses Logs?

    2. Is this always true whenever a limit of a polynomial which is raised to a fraction whose denominator approaches to zero?

    3. Natural Logarithms can also be?

    4. And what if for example a limit approaches to either POSITIVE of NEGATIVE Infinity, how could it be substituted to the function given, what would be the answer and how should it be done?

    THANKS AGAIN IN ADVANCE!!!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    There is no reason not to take that "4" out of the exponent along with the "1/x":

    ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}
    and, by L'Hopital's rule
    ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}
    which is 0.
    Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
    So the final answer is 0?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    There is no reason not to take that "4" out of the exponent along with the "1/x":

    ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}
    and, by L'Hopital's rule
    ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}
    which is 0.
    Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
    So the final answer is 0?
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  8. #8
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    Quote Originally Posted by cutiemike1 View Post
    Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
    So the final answer is 0?
    Hi cutiemike1,

    The natural logarithm is the logarithm to the base "e".
    HallsofIvy showed a better way to reach ln(L).
    Finally, we evaluate L.

    ln(L)=\log_e(L)=0\Rightarrow\ L=e^0=1
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  9. #9
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    Quote Originally Posted by cutiemike1 View Post
    I thought of that one but on my part I used natural logarithm. When I was solving it using natural logarithm I suddenly didn't know what must be done. This one will help me a lot in my final exam. MANY THANKS TO YOU!!
    Now my questions are:

    1. What might be the other cases which uses Logs? typically in cases where the variable is in the exponent

    Logarithm - Wikipedia, the free encyclopedia

    2. Is this always true whenever a limit of a polynomial which is raised to a fraction whose denominator approaches to zero?

    L'Hopital's rule is used in cases where both numerator and denominator
    of a quotient are both approaching 0 or both approaching infinity.

    To apply that rule, both numerator and denominator are differentiated.


    3. Natural Logarithms can also be? The natural logarithm is defined as.... lnx=\log_ex

    4. And what if for example a limit approaches to either POSITIVE of NEGATIVE Infinity, how could it be substituted to the function given, what would be the answer and how should it be done?

    THANKS AGAIN IN ADVANCE!!!
    I don't have a suitable answer to (4.) above.
    Hopefully you will receive a reply, but I'd advise maybe starting a new thread,
    quoting some specific examples and how far you can get with them.
    You'd get some good responses then as logs are not my forte!
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  10. #10
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    I understood the whole thing now. Thanks a lot to the both of you.
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