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I have tried to solve limit, which wolfram says that DNE, but according to my calculations it is equal to 0. Limit is given below

$$\begin{align} \lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi) &=\sin(\sqrt{n^2+1}\pi-n\pi+n\pi) \\ &=(-1)^n\sin(\sqrt{n^2+1}\pi-n\pi) \\ &=(-1)^n\sin\left(\frac{(\sqrt{n^2+1}\pi-n\pi)(\sqrt{n^2+1}\pi+n\pi)}{(\sqrt{n^2+1}\pi+n\pi)}\right) \\ &=(-1)^n\sin\left(\frac{n^2\pi^2+\pi^2-n^2\pi^2}{\sqrt{n^2+1}\pi+n\pi}\right) \\ &=(-1)^n\sin\left(\frac{\pi^2}{n\pi(\sqrt{1+\frac{1}{n^2}}+1}\right) \\ &=0 \end{align}$$

It is because denominator of sin goes to infinity so everything inside sin goes to 0. as we know, sin of that would go to 0 too. And we know that $(-1)^n$ is bounded so we got that bounded * 0 has to be equal to 0. Am I doing some mistake here ?

Blue
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paweta
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3 Answers3

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I don't think you need such a long computation. Once you write your sequence as $$ a_n=(-1)^n\sin((\sqrt{n^2+1}-n)\pi) $$ you can just observe that $\lim\,(\sqrt{n^2+1}-n)=0$ and, by continuity, $$ \lim\, a_n=0. $$

GReyes
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The step $\sin(\sqrt{n^2+1}\pi) = (-1)^n\sin(\sqrt{n^2+1}\pi-n\pi)$ holds if $n$ is an integer but does not hold if $n$ is not an integer.

Let's look at this another way.

$\sqrt{n^2+1}$ is unbounded and increasing. When $n$ is large $\sqrt{n^2+1}\approx n.$

Suppose $n\in \mathbb R$

For large values of $n$ there will be values of $\sin(\sqrt{n^2+1}\pi)$ that equal $1, -1$ and $0$ and the limit does not exist.

e.g. $\sin (\sqrt{100.5^2 + 1}\pi) = 0.9999$

If $n\in \mathbb N$ then for large values of $n$

$\sin(\sqrt{n^2+1}\pi)\approx \sin n\pi = 0$

user317176
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The issue seems to be how $n$ is defined. Ordinarily we treat it as an integer variable, typically using a different letter like $x$ or $t$ if we mean a real variable. Then with $n$ an integer variable, the limit is zero as various answers imply. But if WA treats $n$ as a real variable, it will try to evaluate the "limit" $\sin(n\pi)$ for all real $n$ and, of course, this fails.

I do not know enough about WA to describe how to make it recognize $n$ as an integer variable, but I suppose that a brute-force way would be to put in $\lfloor{n}\rfloor$.

Oscar Lanzi
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