- #1

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- 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