limit of $\lim_{n\to \infty} (-1)^{n-1} \sin(\pi \sqrt{n^2 + 0.5n + 1})$ what am I doing wrong?

Solution 1:

The problem is that as $n\to\infty$, $\sqrt{n^2 +0.5n + 1}$ diverges, and $\sqrt{n^2 +0.5n + 1} -n \neq 0$. We have \begin{align*} \lim_{n\to\infty} \sqrt{n^2 +0.5n + 1} -n &= \lim_{n\to\infty} \sqrt{(n + 0.25)^2 + 1 - 0.25^2} - n\\ &= \lim_{n\to\infty} \frac{(\sqrt{n^2 +0.5n + 1} - n)(\sqrt{n^2 +0.5n + 1} + n)}{\sqrt{n^2 +0.5n + 1} + n}\\ &= \lim_{n\to\infty} \frac{0.5n + 1}{\sqrt{n^2 +0.5n + 1} + n} = \frac{1}{4} \end{align*}

Solution 2:

hint

Begin by noting that

$$(-1)^{n-1}\sin(n\pi +X)=-\sin(X)$$

then, use the fact that when $ x\to 0$,

$$\sqrt{1+x}=1+\frac x2(1+\epsilon(x))$$ with $\lim_{x\to 0}\epsilon(x)=0$.

So

$$\sqrt{1+\frac{1}{2n}+\frac{1}{n^2}}=1+\frac{1}{4n}(1+\epsilon(n))$$ with $\lim_{n\to+\infty}\epsilon(n)=0$.