MHB How to Solve This Trigonometric Limit as x Approaches Infinity?

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The limit of the function $$\lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5})$$ does not exist because as x approaches infinity, the expression inside the sine function tends to infinity, causing the sine to oscillate between -1 and 1. The discussion highlights that any attempt to introduce additional terms does not affect the limit, as they become negligible compared to the dominant cubic term. Ultimately, the oscillatory nature of the sine function prevents the limit from converging to a specific value. Therefore, the conclusion remains that the limit does not exist.
Francolino
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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:

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

Thank you so much for your time! :)
 
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Francolino said:
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:

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

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.
 
Thanks again, for answering. :)

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

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

What do you think?
 
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Francolino said:
Thanks again, for answering. :)

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

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

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:

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

so your n argument doesn't actually change anything.

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

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