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- Dec 16th 2006, 10:56 PM #1

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- Dec 17th 2006, 12:33 AM #2

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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

- Dec 17th 2006, 01:06 AM #3

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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

- Dec 17th 2006, 06:19 AM #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.