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  1. #1
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    Question please help...



    thanks.
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  2. #2
    Grand Panjandrum
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    Prove that between two zeros of $\displaystyle e^x\, \sin(x)=1$ there is a zeros of $\displaystyle e^x\,\cos(x)=-1$

    Let: $\displaystyle f(x) = e^x\, \sin(x) - 1$

    First observe that:

    $\displaystyle e^x\,\cos(x)+1=f'(x)-f(x)$,

    Now if $\displaystyle x_1$ and $\displaystyle x_2$ are consecutive zeros of $\displaystyle f(x)$ then the signs of $\displaystyle f'(x_1)$ and $\displaystyle f'(x_2)$ would be different (the other possibility that one or both of these is zero does not occur as that would imply that for one ,or both, of the zeros that $\displaystyle \cos(x)=\sin(x) \Rightarrow x=\pi/4+n\pi$, but such a point cannot be a zero of $\displaystyle f(x)$).

    But the sign of $\displaystyle f'(x_1)$ and $\displaystyle f'(x_2)$ being different, implies that $\displaystyle e^x\,\cos(x)+1$ is positive at one root and negative at the other, so as $\displaystyle e^x\,\cos(x)+1$ is continuous, by the intermediate value theorem must be zero at some point between $\displaystyle x_1$ and $\displaystyle x_2$. which proves the result.

    RonL
    Last edited by CaptainBlack; Dec 17th 2006 at 06:40 AM.
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  3. #3
    Grand Panjandrum
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    Find where the function:

    $\displaystyle f(x)=\left( x+\frac{1}{x} \right) ^x $

    is an increasing and where it is a decreasing function.

    First we restrict attention to $\displaystyle x>0$, as $\displaystyle f(x)$ is complex or undefined otherwise.

    Rewrite $\displaystyle f(x)$ as: $\displaystyle f(x)=e^{x\,\ln(x+1/x)}$

    Now $\displaystyle f(x)$ is differentiable on $\displaystyle (0,\, \infty)$and its derivative is:

    $\displaystyle \frac{df}{dx}=\left[\frac{x^2}{x^2+1}+\ln(x+1/x) \right] e^{x\,\ln(x+1/x)}$.

    But the term in the square brackets is always positive, as is the term with the exponential. Therefore $\displaystyle f(x)$ is increasing on $\displaystyle (0, \infty)$.

    RonL
    Last edited by CaptainBlack; Dec 17th 2006 at 01:17 AM.
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  4. #4
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    First you probably mean,
    $\displaystyle \lim_{x\to 0^+}x^{x^x}$
    You can write,
    $\displaystyle x^{\exp(x\ln x)}$
    Thus,
    $\displaystyle \exp(\ln x\exp(x\ln x))$
    Let us look at the term inside,
    $\displaystyle \ln x\exp(x\ln x)$
    Now,
    $\displaystyle \lim_{x\to 0^+}\ln x=-\infty$
    And,
    $\displaystyle \lim_{x\to 0^+}\exp (x\ln x)=1$
    By using the L'Hopital rule inside and then applying the limit composite rule.
    Thus,
    $\displaystyle \lim_{x\to 0^+}\ln x\exp (x\ln x)=-\infty$ by the extended reals.

    Thus,*)
    $\displaystyle \lim_{x\to -\infty}\exp x=0$
    If we subtract 1 from this limit we have -1 as the answer to the limit question.


    *)I am note sure about this. Because I employed the limit composite rule for infinite limits (limits that do not exists). Thus, it might not work.
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