My Math Forum Sandwich Theorem Trig Funcs.

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 July 8th, 2019, 05:44 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 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.
 July 8th, 2019, 06:13 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573 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 ... Thanks from babaliaris
July 8th, 2019, 06:18 AM   #3
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Joined: Oct 2015
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Quote:
 Originally Posted by skeeter 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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