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I have to find this limit:

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

My attempt:

Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1} \end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\ \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0 \end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

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    $\begingroup$ To make the argument a bit easier, you can use the fact that $\lim_{n\to\infty}|a_n|=0$ implies $\lim_{n\to\infty} a_n=0$. This will eliminate the $(-1)^n$ term entirely. $\endgroup$
    – Clayton
    Commented Jan 5, 2021 at 20:22

5 Answers 5

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The first inequality is not true in general: if $a\leqslant b$ and $c<0$, then $ac\color{red}{\geqslant}bc$.

But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $0$ indeed.

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Fact 1. If $a_n\leq b_n$ for large $n$ then $\displaystyle\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n$, provided the limits exist.

Fact 2. If $\displaystyle\lim_{n\to\infty}|a_n|=0$ then $\displaystyle\lim_{n\to\infty}a_n=0$.

From fact 1 we have $$ 0\leq\lim_{n\to\infty}\left|\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}\right| \leq\lim_{n\to\infty}\frac{\sqrt{n}}{n+1} =0 $$ and so we can apply fact 2 to obtain $$ \lim_{n\to\infty}\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1} =0. $$

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\begin{align} \lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

This is one of those problems where you are supposed to take a step back and make things easy on yourself.

Let

$$f(n) = (-1)^{n}\sin(n^{n})$$

and let

$$g(n) = \frac{\sqrt{n}}{n+1} < \frac{1}{\sqrt{n}}.$$

You want

$$\lim_{n\to \infty} [f(n) \times g(n)].$$

As $n \to \infty, f(n)$ is a bounded function.
That is $~-1 \leq f(n) \leq +1~$ for all $n$.

Also, clearly, as $n \to \infty, g(n) \to 0.$

Therefore, as $n \to \infty, \{f(n) \times g(n)\} \to 0.$

Edit
Adding some formality:
For all $n, |f(n) \times g(n)| \leq |g(n)|$ which goes go zero.

Therefore, since $|f(n) \times g(n)| \to 0,~ \{f(n) \times g(n)\} \to 0.$

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    $\begingroup$ My English it is very poor (I have written limited) :-( $\endgroup$
    – Sebastiano
    Commented Jan 5, 2021 at 20:37
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    $\begingroup$ @Sebastiano +1 for your answer, which I regard as valid. $\endgroup$ Commented Jan 5, 2021 at 20:40
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    $\begingroup$ Ahahah :-) there is not problem for the vote for me :-). However thank you. My version is the one I do in high school. Unfortunately, when I have to do with the upper-bound, the students don't understand what it means because they haven't understood that inequalities are needed. $\endgroup$
    – Sebastiano
    Commented Jan 5, 2021 at 20:44
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If $(a_n)$ is a bounded sequence and $(b_n)$ is an infinitesimal sequence, then the product sequence $(a_n \cdot b_n)$ is also infinitesimal.

Starting from your sequence,

$$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$$

Being $\forall n\in \mathbb{N}$ then $-1 \leq (-1)^{n} \le 1$ and $-1\leq \sin (n^n)\leq 1$ two bounded sequences then $$\lim_{n \rightarrow \infty} \frac{(-1)^{n}}{n+1}=\lim_{n \rightarrow \infty} \frac{\sin (n^n)}{n+1}=0$$

$$\lim_{n \rightarrow \infty} \underbrace{[(-1)^{n}\sin(n^{n})]}_{\text{bounded sequences}}\,\frac{\sqrt{n}}{n+1}=\lim_{n \rightarrow \infty} [(-1)^{n}\sin(n^{n})]\cdot \underbrace{\sqrt{\frac{n}{(n+1)^2}}}_{\text{infinitesimal sequence}}=0$$

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You're either right or very nearly right: in either case I would be more comfortable putting some absolute value signs on the function to get rid of the $(-1)^n$ terms.

Here's why: let's say we have some $f(x)$ such that $\lim_{x \to a} |f(x)| = 0$. Notice by the linearity of limits that $\lim_{x \to a} -|f(x)| = -\lim_{x \to a} |f(x)| = 0$.

So, because for any $f(x)$ we have $-|f(x)| \leq f(x) \leq |f(x)|$, by the squeeze theorem we have that $\lim_{x \to a} f(x) = 0$.

So, the $(-1)^n$ terms aren't an issue, but I would state this explicitly in your work.

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