 #1
songoku
 2,090
 277
 Homework Statement:

Find
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
 Relevant Equations:
 Limit
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi (\sqrt{n^2+n}n))$$
$$=\lim_{n \rightarrow \infty} \sin^{2} \left(\pi \left((\sqrt{n^2+n}n) . \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \lim_{n \rightarrow \infty} \left(\frac{n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \left(\frac{1}{2}\right)\right)$$
$$=1$$
But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.
Where is the mistake in my reasoning?
Thanks
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi (\sqrt{n^2+n}n))$$
$$=\lim_{n \rightarrow \infty} \sin^{2} \left(\pi \left((\sqrt{n^2+n}n) . \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \lim_{n \rightarrow \infty} \left(\frac{n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \left(\frac{1}{2}\right)\right)$$
$$=1$$
But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.
Where is the mistake in my reasoning?
Thanks