Evaluate the limit of the sequence: $a_n = \sin \left( 2\pi \sqrt{n^2+n} \right)$.
Solution
We are using the transfer principle. We substract a $2\pi n$ and we have that:
$$\sin \left ( 2\pi \sqrt{n^2+n} \right )= \sin \left ( 2\pi \sqrt{n^2+n}-2\pi n \right ) = \sin \left ( \frac{2\pi}{\sqrt{1+ \frac{1}{n}}+1} \right )$$
However $\displaystyle \frac{2\pi}{\sqrt{1+ \frac{1}{n}}+1}\rightarrow \pi$. Hence the limit of the sequence is $0$.
A plot that confims the above.
The exercise can also be found in mathematica.gr
Solution
We are using the transfer principle. We substract a $2\pi n$ and we have that:
$$\sin \left ( 2\pi \sqrt{n^2+n} \right )= \sin \left ( 2\pi \sqrt{n^2+n}-2\pi n \right ) = \sin \left ( \frac{2\pi}{\sqrt{1+ \frac{1}{n}}+1} \right )$$
However $\displaystyle \frac{2\pi}{\sqrt{1+ \frac{1}{n}}+1}\rightarrow \pi$. Hence the limit of the sequence is $0$.
The exercise can also be found in mathematica.gr
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