How can I formally prove the following function is bounded? I have recently started learning Calculus, and upon stumbling on the following question, I got a little stuck. 
The question:
Determine if the function $f:\mathbb R\to\mathbb R$ given by $f(x) = {x\sin(x)\over |x| + 1}$ is bounded and provide a justification (a proof).
Since I know that the denominator is never going to be a negative number, I have tried to get rid of the absolute signs $|x| + 1 ≥ 1$  and solve for x like so: $-1≥x+1≥1$
$-2≥x≥0$
I thought that by doing so, I will be able to get some information about the numerator, however, I just confused myself. 
Would anybody be kind enough to set me on the right direction?
 A: 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 
A: Thanks for the great hints guys! I have tried proving this statement like so:
${x\sin(x)\over |x| + 1}$ 
$|{x\sin(x)\over |x| + 1|}|$

${|x\sin(x)|\over |x| + 1|}$
${|x|*|sin(x)|\over |x| + 1|}$
Since ${|xsin(x)|≤1}$ we can multiply the inequality $xsin(x)≤1$ by ${|x|\over |x| + 1}$ on both sides which will produce an expression that is still ≤1.
like so:

${|x|*1\over |x| + 1}≤1$ 
${|x|\over |x| + 1}≤1$ 
Is this sufficient to prove the statement is bounded?
