Question #5fa6a

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Question #5fa6a

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Answer 1 · 2016-06-01T18:08:34

$\frac{2 + 2 x + sin 2 x}{(2 x + sin 2 x) e^{sin x}} = \frac{2 + 2 x + sin 2 x}{(2 x + sin 2 x)} \cdot \frac{1}{e^{sin x}}$ $(2+2x+sin2x)/((2x+sin2x)e^(sinx)) = (2+2x+sin2x)/((2x+sin2x))*1/e^sinx$

$= [\frac{2}{2 x + sin 2 x} + \frac{2 x + sin 2 x}{2 x + sin 2 x}] \frac{1}{e^{sin x}}$ $= [2/(2x+sin2x) +(2x+sin2x)/(2x+sin2x)]1/e^sinx$

$= [\frac{2}{2 x + sin 2 x} + 1] \frac{1}{e^{sin x}}$ $= [2/(2x+sin2x) +1]1/e^sinx$

As $x \to \infty$ $xrarroo$ , we see that

$\frac{2}{2 x + sin 2 x} \to 0$ $2/(2x+sin2x) rarr 0$ , so

$[\frac{2}{2 x + sin 2 x} + 1] \to [0 + 1] = 1$ $[2/(2x+sin2x) +1] rarr [0+1]=1$ .

Furthermore, as $x \to \infty$ $xrarroo$ , the factor

$\frac{1}{e^{sin x}}$ $1/e^sinx$ oscillates between $e$ $e$ and $\frac{1}{e}$ $1/e$

Therefore, the function

$f (x) = \frac{2 + 2 x + sin 2 x}{(2 x + sin 2 x) e^{sin x}}$ $f(x) = (2+2x+sin2x)/((2x+sin2x)e^(sinx))$ does not approach a limit as $x \to \infty$ $xrarroo$ .

As $x \to \infty$ $xrarroo$ the relative maximum values of $f$ $f$ approach $e$ $e$ and the minimum values approach $\frac{1}{e}$ $1/e$ .

In the graph below, you see the function and the line $y = e$ $y=e$ . (When I also included $y = \frac{1}{e}$ $y=1/e$ the graph developed holes in it.)

You can (and should) scroll in or out and drag the graph around to see what happens. (When you leave the page and return the default view will be there again.)

graph{(y- (2+2x+sin2x)/((2x+sin2x)e^(sinx)) )(y-e) = 0 [-1.55, 30.5, -5.19, 10.83]}

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Question #5fa6a

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