# 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

• I guess you need a definition of "limit as $x$ approaches infinty" when the function is undefined for a sequence of values going to infinity. – GEdgar Oct 22 '17 at 18:32
• i think the Limit doesn't exist – Dr. Sonnhard Graubner Oct 22 '17 at 18:33
• that limit does not exists. takes $x= n\pi$ and change with pi/2 you will to different limits – Guy Fsone Oct 22 '17 at 18:36
• You don't need series. The set of limit values of $\tan x$ is $(-\infty,+\infty)$ so there is no limit – Raffaele Oct 22 '17 at 19:02
• The book where I got this question had a finite answer but not a detailed solution – user481779 Oct 22 '17 at 21:39

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

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.

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.

$\lim_{x\to+\infty}\frac{\sin x\tan x }{x^3+x^2}$

$\sin x\tan x =\dfrac{\sin^2x}{\cos x} \to \infty$ 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.