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limit of function ("sandwich" method)
Using the "sandwich" method prove that [tex]\lim_{n\rightarrow \propto }(\frac{sin(n)}{n})=0[/tex]
[tex]x_n \leq y_n \leq z_n[/tex]
[tex]\lim_{n\rightarrow \propto }(x_n) \leq \lim_{n\rightarrow \propto }(y_n) \leq \lim_{n\rightarrow \propto }(z_n)[/tex]
I am honestly little bit confused at this point.
If the answer is:
[tex]\frac{-1}{n} \leq \frac{sin(n)}{n} \leq \frac{1}{n}[/tex]
then my question is if [itex]n=-\frac{\pi}{4}[/itex] then [tex]\frac{-1}{-0.785}[/tex] will be not less or equal to [tex]\frac{\sqrt{2}}{2*(-0.785)}[/tex], where -0.785=[itex]-\frac{\pi}{4}[/itex], where [itex]\pi \approx 3.14[/itex].
Thanks in advance.
Homework Statement
Using the "sandwich" method prove that [tex]\lim_{n\rightarrow \propto }(\frac{sin(n)}{n})=0[/tex]
Homework Equations
[tex]x_n \leq y_n \leq z_n[/tex]
[tex]\lim_{n\rightarrow \propto }(x_n) \leq \lim_{n\rightarrow \propto }(y_n) \leq \lim_{n\rightarrow \propto }(z_n)[/tex]
The Attempt at a Solution
I am honestly little bit confused at this point.
If the answer is:
[tex]\frac{-1}{n} \leq \frac{sin(n)}{n} \leq \frac{1}{n}[/tex]
then my question is if [itex]n=-\frac{\pi}{4}[/itex] then [tex]\frac{-1}{-0.785}[/tex] will be not less or equal to [tex]\frac{\sqrt{2}}{2*(-0.785)}[/tex], where -0.785=[itex]-\frac{\pi}{4}[/itex], where [itex]\pi \approx 3.14[/itex].
Thanks in advance.
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