Finding the values of $x$ which satisfy $\lim_{n\to \infty} \left|\frac{-(x+3)\sqrt{n+1}}{2\sqrt{n}}\right|<1$ $$\lim_{n\to \infty} \left|\frac{-(x+3)\sqrt{n+1}}{2\sqrt{n}}\right|<1$$
$$\lim_{n\to \infty} \left|\frac{-(x+3)(n^\frac{1}{2}+\cdots)}{2n^\frac{1}{2}}\right|<1$$
Divide by $n^\frac{1}{2}$ gives: 
$$\left|\frac{-(x+3)}{2}\right|<1$$
Is this step part correct? I know how to finish this off, if this part is correct.
 A: From what you write it is not clear what happens in your manipulation. A better way would be:
If $x \in \mathbb{R}$, then
$$
\lim_{n \to \infty}\bigg| \frac{-(x+3)\sqrt{n+1}}{2\sqrt{n}} \bigg| = | -(x+3)| \bigg( \lim_{n \to \infty}\frac{\sqrt{1 + \frac{1}{n}}}{2} \bigg) = \bigg| \frac{-(x+3)}{2} \bigg|.
$$
A: Notice, $$\lim_{n\to \infty}\left|\frac{-(x+3)\sqrt{n+1}}{2\sqrt n}\right|<1$$
$$\lim_{n\to \infty}\left|\frac{-(x+3)\sqrt n\sqrt{1+\frac{1}{n}}}{2\sqrt n}\right|<1$$
$$\lim_{n\to \infty}\left|\frac{-(x+3)\sqrt{1+\frac{1}{n}}}{2}\right|<1$$
$$\left|\frac{-(x+3)\sqrt{1+0}}{2}\right|<1$$
$$\left|\frac{-(x+3)}{2}\right|<1$$
$$\frac{|x+3|}{2}<1$$
$$ |x+3|<2$$ $$\iff-2<x+3<2$$
$$\iff \color{red}{-5<x<-1}$$ 
A: $lim_{n\rightarrow \infty} \frac{-(x+3)\sqrt{n+1}}{2\sqrt{n}}=lim_{n\rightarrow \infty} \frac{-(x+3)}{2}\cdot lim_{n\rightarrow \infty} \frac{\sqrt{n+1}}{\sqrt{n}}$ if both limits exist. Then if you agree that $lim_{n\rightarrow \infty} \frac{\sqrt{n+1}}{\sqrt{n}}=1$ your solution is good.
