# What are the limits of these sine functions as x approaches infinity?

I have two different limit sine functions:

$$\lim_{x\to \infty} {\sin x\over x}$$ $$\lim_{x\to \infty}x^2\sin\left({3\over x^2}\right)$$

My thoughts are that as $x$ becomes infinitely large, then the function would become very small, close to $0$. And because sine of $0$ is $0$, the limit would equal $0$.

Is this correct?

HINT :

For the first, note that we have $|\sin x|\le 1$.

For the second, we have$$x^2\sin\left(\frac{3}{x^2}\right)=\frac{\sin\left(\dfrac{3}{x^2}\right)}{\dfrac{1}{x^2}}=\frac{\sin\left(\dfrac{3}{x^2}\right)}{\dfrac{3}{x^2}}\cdot 3$$

$\sin x$ is bounded, $1/x$ is infinitesimal, then it is well known that the product of two such functions is infinitesimal, hence tends to $0$.

For the other one: we know that $\sin \xi\sim \xi$ as $\xi$ approaches to $0$, thus $\sin(3/x^2)\sim3/x^2$, thus the second limit is $$\lim_{x\to\infty}x^2\sin(3/x^2)=\lim_{x\to\infty}x^2(3/x^2)=3\;\;.$$

For the first limit, we know that $\sin x$ oscillates in between $-1$ and $1$. That is, whatever value of $x$ you put into the sine function, you'll get a number between $-1$ and $1$, inclusive. As $x$ increases, the denominator becomes increasing large and dominates the value of $\sin x$. So the entire expression tends to $0$ as $x$ gets very large simply because you have $\frac{[-1,1]}{\mathrm{huge \, number}}$.

For the second, make use of the fact that $\displaystyle \sin \left(\frac{3}{x^2}\right)$ is about $\dfrac{3}{x^2}$ when $x \to \infty$. The reason this is true is because as $a \to 0$ we have $\sin a \to a$. And $\dfrac{3}{x^2}$ does tend to $0$ as $x \to \infty$. (You can see this from a basic Taylor expansion of the sine function)

Edit: This is meant to be more of a intuitive thing and nothing formal!

The limit of $\dfrac{\sin(x)}{x}$ as $x \to \infty$ is zero. Clearly the limit of $\sin(x)$ is between 1 and -1. Hence the limit of $\frac{\sin(x)}{x}$ is between $\frac{-1}{x}$ and $\frac{1}{x}$. Therefore, the limit of $\frac{1}{x}$ as $x \to \infty$ is zero, hence limit of $\dfrac{\sin(x)}{x}$ as $x \to \infty$ is zero. A similar argument can be applied to the second problem. The limit of $x^2 \sin\left(\frac{3}{x^2}\right)$ as $x \to \infty$ is $3$.

• any reason for the downvote? I've only tried to answer the question and its really not fair downvoting for no reason. – John_dydx Jun 27 '15 at 17:59
• (Not my downvote but) at this point your answer kind of just looks like a talking Wolfram Alpha. Downvoter probably found it not very useful to the OP. – miradulo Jun 27 '15 at 18:01
• I can obviously explain more if needed but thanks for pointing out. – John_dydx Jun 27 '15 at 18:02

In the second part, try obtaining an expression of the form: $$\lim_{x \to 0} \dfrac{\sin a}{a}$$

With Asymptotic analysis:

$\dfrac{\sin x}x =_{\infty} O\Bigl(\dfrac1x\Bigr)$, hence $\;\dfrac{\sin x}x\xrightarrow[x\to\infty]{} 0$

$\sin\dfrac3{x^2}\sim_{\infty}\dfrac3{x^2}$, hence $\;x^2\sin\dfrac3{x^2}\sim_{\infty}3$, thus $\;x^2\sin\dfrac3{x^2}\xrightarrow[x\to\infty]{}3$.