How to evaluate $\lim_{x\to+\infty}\frac{\sin x\tan x }{x^3+x^2} $? 
Evaluate
  $$\lim_{x\to+\infty}\frac{\sin x\tan x }{x^3+x^2}
$$


I have tried using series but couldn't solve the problem
 A: Let $y=\frac1x$. Then we have $x=\frac1y$,  $\lim_{y\to0}\frac{\sin \frac1y\tan \frac 1y }{(\frac1y)^3+(\frac1y)^2}$. Now let's show that $$\lim_{y\to0}\sin \frac1y$$ does not exist. We will write:
$${Y_{n1}}=\frac1{n*\pi} = \frac1\pi,\frac1{2\pi},\frac1{3\pi},\frac1{3\pi},\frac1{4\pi},....-->0; sin(\frac1{Y_{n1}})=0 (always);\lim_{y\to0}\sin \frac1{Y_{n1}}=0$$
$${Y_{n2}}=\frac2{(4n+1)*\pi} = \frac2{5\pi},\frac2{9\pi},\frac2{13\pi},....-->0; sin(\frac1{Y_{n2}})=1(always);\lim_{y\to0}\sin \frac1{Y_{n2}}=1$$
$${Y_{n3}}=\frac2{(4n-1)*\pi} = \frac2{3\pi},\frac2{7\pi},\frac2{11\pi},....-->0; sin(\frac1{Y_{n3}})=-1(always);\lim_{y\to0}\sin \frac1{Y_{n2}}=-1$$
So, in one pointwe have 3 different limits. But it is not possible. It means that limit does not exist
A: Consider the equation
$$
\sin x\tan x=x^k
$$
for $k$ a positive integer. It's not difficult to show that each of these equations admits an increasing sequence of solutions $(x_{k,n})$ so that
$$
\lim_{n\to\infty}x_{k,n}=\infty
$$
Since
$$
\lim_{n\to\infty}\frac{\sin x_{2,n}\tan x_{2,n}}{x_{2,n}^3+x_{2,n}^2}=
\lim_{n\to\infty}\frac{1}{x_{2,n}+1}=0
$$
whereas
$$
\lim_{n\to\infty}\frac{\sin x_{3,n}\tan x_{3,n}}{x_{3,n}^3+x_{3,n}^2}=
\lim_{n\to\infty}\frac{1}{1+\frac{1}{x_{3,n}}}=1
$$
we can conclude that the limit doesn't exist.
A: The arguments given are of intuitive nature.
The limit $\displaystyle \lim_{x \rightarrow  \infty} \sin{x}$ does not exist but $\sin{x}$  is a bounded function. 
So for the limit $\displaystyle \lim_{x \rightarrow  \infty} \frac{\sin{x}}{x}$  we can use Squeeze Theorem and prove that it is $0$. Look at the Graph.
Sin[x]/x
The limit $\displaystyle \lim_{x \rightarrow  \infty} \tan{x}$ does not exist but $\tan{x}$  is an unbounded function.
So the limit $\displaystyle \lim_{x \rightarrow  \infty} \frac{\tan{x}}{x}$ Does not exist. Also look at another post link. Here is the Graph:
Tan[x]/x
Clearly the function $\displaystyle \frac{\sin{x}\tan{x}}{x^3+x^2}$  is an unbounded function. Here is the Graph:
(Sin[x]Tan[x])/(x^3 + x^2)
So intuitively $\displaystyle \lim_{x \rightarrow  \infty} \frac{\sin{x}\tan{x}}{x^3+x^2}$ Does not exist.
A: $\lim_{x\to+\infty}\frac{\sin x\tan x }{x^3+x^2}
$
$\sin x\tan x 
=\dfrac{\sin^2x}{\cos x}
\to \infty
$
when
$x\to 2\pi n+\pi/2
$
and
$\sin x\tan x 
=\dfrac{\sin^2x}{\cos x}
\to 0
$
when
$x\to 2\pi n
$
so the limit does
not exist.
