Looking at the various comments by OP it seems that the textbooks he is following are not of good quality and instead of providing a step by step approach to limits they are trying to teach a bag of tricks.
For most of the usual limit problems the basic rules of limits and certain standard limits are sufficient:
1) $\displaystyle \lim_{x \to a}f(x) \pm g(x) = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$
2) $\displaystyle \lim_{x \to a}f(x) \cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$
3) $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \dfrac{{\displaystyle \lim_{x \to a}f(x)}}{{\displaystyle \lim_{x \to a}g(x)}}$ provided that $\lim_{x \to a}g(x) \neq 0$.
The above rules apply provided that both limits in RHS exist. Certain standard limits are as follows: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\, \lim_{x \to 0}\frac{\sin x}{x} = 1,\, \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$
Next we come to the particular question here $$\lim_{h \to 0}\frac{5}{\sqrt{5h + 1} + 1}$$ We don't need to think that this is a rational function or not, but just notice that it is an expression of the form $f(h)/g(h)$ where $f, g$ are some functions and hence the rule 3) should be applied. Clearly for the numerator $f(h) = 5$ we see that $\lim_{h \to 0} 5 = 5$ and we need to see if the limit of denominator $g(h) = \sqrt{5h + 1} + 1$ exists and is non-zero. If we see the form of $g(h)$ it looks like sum of two functions and hence rule 1) can be applied. Thus we can write
$\displaystyle \begin{aligned}\lim_{h \to 0}\frac{5}{\sqrt{5h + 1} + 1} &= \dfrac{{\displaystyle \lim_{h \to 0}5}}{{\displaystyle \lim_{h \to 0}\{\sqrt{5h + 1} + 1\}}}\\
&= \dfrac{5}{{\displaystyle \lim_{h \to 0}\{\sqrt{5h + 1} + 1\}}}\\
&= \dfrac{5}{{\displaystyle \lim_{h \to 0}\sqrt{5h + 1} + \lim_{h \to 0}1}}\\
&= \frac{5}{1 + 1} = \frac{5}{2}\end{aligned}$
Just to provide a contrasting example we try to calculate $$\lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1}$$ If we follow as before we will see that limit of both numerator and denominator is $0$ and hence rule 3) can't be applied precisely because the denominator limit is $0$. It is then time to do some manipulation so that the given function can be represented in a form which avoids denominator limit $0$. Such a manipulation is done under the valid assumption that $h \neq 0$. Then
$\displaystyle \begin{aligned}\lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1} &= \lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1}\cdot\frac{\sqrt{5h + 1} + 1}{\sqrt{5h + 1} + 1}\\
&= \lim_{h \to 0}\frac{h\{\sqrt{5h + 1} + 1\}}{5h}\\
&= \lim_{h \to 0}\frac{\sqrt{5h + 1} + 1}{5}\\
&\text{(following as in previous example)}\\
&= \frac{2}{5}\end{aligned}$
As seen in various limit question on this website, most beginners in calculus try to use concepts like continuity, derivative, L'Hospital and even series expansions to solve simple limit problems (most answers given here also try to use these techniques). It is very unfortunate that beginners underestimate the power of simple rules of limits (mentioned above) and jump on high level concepts. IMHO a beginner studying limits for the first time is better off if he is totally unaware of these high level concepts (which are ultimately derived from simpler concept of limits) and instead focuses on the rules of limits.
You need a bit more. That the limit of the sequence is $2$ isn't enough. You also need, that this sequence is monotonous. Then you can say that $2$ is the upper bound.
You can try it this way:
As $n$ approaches infinity, you see that $n$ gets much bigger than 1, so you can neglect it and you get
$$ \lim_{n\rightarrow\infty} \frac{2n}{1+n} = \lim_{n\rightarrow\infty}\frac{2n}{n} = 2. $$
Because $a_n$ is monotonous, you see, that this is indeed the upper bound.
Best Answer
Everything seems to check out. What you essentially have done is this:
since we want to show $\frac{x*\sin(x)}{|x|+1}$ is bounded we take its absolute value so we have
|$\frac{x*\sin(x)}{|x|+1}$| = $\frac{|x*\sin(x)|}{||x|+1|}$ = $\frac{|x|*\|sin(x)|}{||x|+1|}$ By what you said about |$\sin(x)$| $\le$ 1 we get that
$\frac{|x|*\|sin(x)|}{||x|+1|}$ $\le$ $\frac{|x|}{||x|+1|}$
Since |x| $\le$ |x| + 1, we get that $\frac{1}{|x|}$ $\ge$ $\frac{1}{|x|+1}$ which shows us that
$\frac{|x|}{||x|+1|}$ $\le$ $\frac{|x|}{|x|}$ = 1
So yes, you have proven that f(x) := $\frac{x*\sin(x)}{|x|+1}$ is bounded