How to Solve This Trigonometric Limit as x Approaches Infinity?

[Francolino]

Hi to everyone. I'm new on here (in fact, this is my really first message). I need some help with the next limit, I hope you can help me:

\(\displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5})\)​

Thank you so much for your time! :)

 

[Euge]

Hi to everyone. I'm new on here (in fact, this is my really first message). I need some help with the next limit, I hope you can help me:

\(\displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5})\)​

Thank you so much for your time! :)

Hi Francolino,

The limit of your function does not exist. To see why, suppose it did have a limit, $L$. Let $y = \pi x\sqrt[3]{x^3 + 3x^2 + 4x - 5}$. Then $y \to \infty$ as $x \to \infty$. So (*) $L = \lim_{y \to \infty} \sin{y}$. However, as $y$ increases without bound, $\sin(y)$ oscillates between -1 and 1. Therefore, (*) cannot hold.

 

[Francolino]

Thanks again, for answering. :)

\(\displaystyle \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + 0) = \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + [x·(x+1)·\pi - x·(x+1)·\pi])\)

As $x·(x+1)·\pi$ is always a pair number (no matter what intenger $x$ is), so:
\(\displaystyle \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} \pi -x·(x+1)·\pi) = \frac {\sqrt {3}}{2}\)

What do you think?

 

[topsquark]

Thanks again, for answering. :)

\(\displaystyle \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + 0) = \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + [n·(n+1)·\pi - n·(n+1)·\pi])\)

As $n·(n+1)·\pi$ is always a pair number (no matter what intenger $n$ is), so:
\(\displaystyle \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} \pi - n·(n+1)·\pi) = \frac {\sqrt {3}}{2}\)

What do you think?

It doesn't change the previous argument. Look at it this way, the limit of the cubic is essentially linear in x (after the cube root) and will swamp any terms in n, unless perhaps you take n to infinity as well. I'll leave that to the professionals. In this case though, your terms in n will not have an effect when x goes to infinity.

Writing in the "intermediate" step:

\(\displaystyle \displaystyle \lim_{x \to \infty } \sin \left ( x \pi \sqrt [3] {x^3+3x^2+4x-5} - n\cdot(n+1)\cdot\pi \right ) \to \lim_{x \to \infty } \sin \left ( x \pi \sqrt [3] {x^3+3x^2+4x-5} \right )\)

so your n argument doesn't actually change anything.

-Dan

 

[Francolino]

I apologize myself. I commited a mistake typing my last message. Now, the error is fixed.

Forget about $ n $. It's $ x $, now. Sorry.