What is the answer of $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$ I found a limit question in my textbook but i could not obtained the result given by answer key.
Let me introduce the question :  $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$
I tried to separate it into two part such that $ \lim_{n\to \infty}{(-1)}^{(n-1)}\cdot\lim_{n\to \infty}\frac{n}{(n+1)!}$
As you see, this partition gave me "$\text{oscillating (undetermined)}$".$0$
However, the answer is zero. What am I missing? Thank you for your helps ...
 A: The rule that $\lim_n (a_n b_n) = \left( \lim_n a_n\right)\left(\lim_n b_n\right)$ is valid only when both of the latter limits exist, meaning both are actual finite numbers (not $\pm\infty$ and not something that oscillates). And $(-1)^n$ oscillates.
But the limit you started with exists.
$$
\frac{-n}{(n+1)!} \le \frac{(-1)^n \cdot n}{(n+1)!} \le \frac n {(n+1)!}
$$
Now observe that
$$
\frac n {(n+1)!} = \frac n {n+1} \cdot \frac 1 {n!}.
$$
And this one you can partition, finding limits of the two factors separately. The first one approaches $1$ and the second approaches $0.$ Thus the sequence whose limit you seek is squeezed between two sequences whose limit is $0.$
A: $\lim_{n\to\infty}{n\over(n+1)!}=\lim_{n\to\infty} \frac{1}{(n+1)(n-1)!}=0\Longrightarrow\lim_{n\to\infty}{(-1)^{n-1}n\over(n+1)!}=0 $
A: Try squeeze.
$0 = \lim_{n \to \infty}\frac{-n}{(n+1)!} \leq \lim_{n \to \infty}\frac{(-1)^{n-1}n}{(n+1)!} \leq \lim_{n \to \infty}\frac{n}{(n+1)!} = 0$
A: You managed to rewrite your sequence as $x_n=y_n\cdot z_n$ where $y_n$ is bounded and $z_n\to0$. (In your case, $y_n=(-1)^n$ and $z_n=n/(n+1)!$.
In general, the following is true: If $y_n$ is bounded and $\lim\limits_{n\to\infty} z_n=0$ then $$\lim\limits_{n\to\infty} y_nz_n=0.$$
Proof. Let $M$ be such that $|y_n|\le M$ for each $n$. If we are given $\newcommand{\ve}{\varepsilon}\ve>0$, then there is $n_0$ such that for $n\ge n_0$ we have
$$|z_n| < \frac\ve{M}.$$
Then for every $n\ge n_0$ we also get
$$|y_nz_n| < M\cdot \frac\ve{M} = \ve.$$
This shows that the sequence $(y_nz_n)$ converges to zero. $\square$
