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July 8th, 2019, 05:44 AM   #1
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Sandwich Theorem Trig Funcs.

For some reason I struggle using the sandwich theorem in solving trigonometric function limits.

Example
1.png

Solution
2.png

But I can not understand how he does that. My try using $\displaystyle -|θ| < sinθ < |θ|$ :
$\displaystyle
-|2x| < sin2x < |2x| \Leftrightarrow \frac{-|2x|}{x} < \frac{sin2x}{x} < \frac{|2x|}{x}
$

what now? how does the books reaches this:
$\displaystyle
\frac{-1}{x} < \frac{sin2x}{x} < \frac{1}{x}
$
????

How does he also get rid of the absolute value?

I solved this with words $\displaystyle \lim_{x \to \infty} \frac{sin2x}{x}$, $\displaystyle \lim_{x \to \infty} sin2x = a \epsilon R$, so $\displaystyle \frac{a}{\lim_{x \to \infty}x}=0$

Last edited by babaliaris; July 8th, 2019 at 05:50 AM.
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July 8th, 2019, 06:13 AM   #2
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the value of sin(any angle) is inclusively between -1 and 1, i.e

$-1 \le \sin(2x) \le 1$

since $x \to \infty$, $x$ is a positive, non-zero number and one can divide each term in the above inequality by $x$

$\dfrac{-1}{x} \le \dfrac{\sin(2x)}{x} \le \dfrac{1}{x}$

as $x \to \infty$, both $-\dfrac{1}{x}$ and $\dfrac{1}{x}$ tend to zero ...
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July 8th, 2019, 06:18 AM   #3
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Quote:
Originally Posted by skeeter View Post
the value of sin(any angle) is inclusively between -1 and 1, i.e

$-1 \le \sin(2x) \le 1$

since $x \to \infty$, $x$ is a positive, non-zero number and one can divide each term in the above inequality by $x$

$\dfrac{-1}{x} \le \dfrac{\sin(2x)}{x} \le \dfrac{1}{x}$

as $x \to \infty$, both $-\dfrac{1}{x}$ and $\dfrac{1}{x}$ tend to zero ...
When can this inequality $\displaystyle -|θ| < sinθ < |θ|$ be useful?
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