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Evaluate the given limit. $$\lim_{x\to\infty}\frac{2x^3\sin{(4x)}}{6x^3+5}$$

My attempt: $$\lim_{x\to\infty}\frac{2x^3\sin{(4x)}}{6x^3+5}=\lim_{x\to\infty}\frac{2\sin{(4x)}}{6}$$

MathFail
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William
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    You should use more parentheses if you intend all of 6x^3+5 to be part of the denominator. a+b/c+d is typically interpreted as $a+\dfrac{b}{c}+d$, not as $\dfrac{a+b}{c+d}$. See here for how to type with MathJax and $\LaTeX$ here. – JMoravitz Jun 01 '23 at 12:35
  • Assuming that was all part of the denominator, then yes... you get to simplify it to a trig function times a constant. You should know that the sine function famously just keeps going back and forth forever, not getting any closer to any particular thing... which would imply that there is no limit. – JMoravitz Jun 01 '23 at 12:36
  • Thank you for the correction. I didn’t know how to write it well. – William Jun 01 '23 at 12:39

2 Answers2

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From where you left of we can write this as $\frac{2}{6}\lim_{x\to\infty}{\sin{4x}}$. You can see here: Prove that $\lim_{x→\infty}\sin(x)$ doesn't exist that $\lim_{x\to\infty}{\sin{x}}$ diverges, thus $\lim_{x\to\infty}{\sin{4x}}$ also diverges, meaning it doesn't exist.

bb_823
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Take $x_n=n\pi$, you get

$$\lim_{n\to\infty}\frac{2(n\pi)^3\sin{(4n\pi)}}{6(n\pi)^3+5}=0$$

Take $\displaystyle x_n=\frac{2n\pi+\frac{\pi}2}4$, you get

$$\lim_{n\to\infty}\frac{2\cdot(2n\pi+\frac{\pi}2)^3\sin(2n\pi+\frac{\pi}2)}{6\cdot(2n\pi+\frac{\pi}2)^3+5}=\frac13$$

So the limit doesn't exist.

MathFail
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