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, nonzero 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 ... 
July 8th, 2019, 06:18 AM  #3  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
 

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funcs, sandwich, theorem, trig 
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