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Limit of $6x\sin(\frac{4}{x})$ as $x$ approaches infinity. I know its $24$ thanks to wolfram alpha. I don't know how to get there. Just need to understand how this is done so I am not lost in the future.

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  • $\begingroup$ Wolfram alpha suggests series expansion. $\endgroup$ Apr 10 '16 at 3:05
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Make a change of variable: ($n = 1/x$) $$\lim_{x\to \infty}6x\sin \frac 4x=\lim_{n\to 0}\frac{6\sin 4n}{n}$$ $$=\lim_{n\to 0}\frac{24\sin 4n}{4n}$$ $$=\lim_{n\to 0}24\left(\frac{\sin 4n}{4n}\right)$$ $$=24$$

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  • $\begingroup$ @pewpew Note that $n=\frac{1}{x}$ $\endgroup$
    – Ian Miller
    Apr 10 '16 at 3:13
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I thought it might be instructive to present a way forward that relies on elementary tools only. To that end, recall from geometry that the sine function satisfies the inequalities

$$x\cos(x)\le \sin(x)\le x$$

for $0\le x\le \pi/2$.

Then, we have

$$24\cos\left(\frac{4}{x}\right)\le 6x\sin\left(\frac{4}{x}\right)\le 24$$

for $x\ge 8/\pi$, whence upon applying the squeeze theorem yields the limit

$$\lim_{x\to \infty}6x\sin\left(\frac{4}{x}\right)=24$$

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$$\lim_{x\to\infty} 6x\sin\left(\frac{4}{x}\right) = \lim_{x\to\infty} 6x\left(\frac{4}{x}+O\left(x^{-3}\right)\right) = \lim_{x\to\infty} 24+O(x^{-2}) = 24$$

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I would try to use a famous limit ($\frac{\sin u}{u}$ where as $u$ tends to $0$ the limit tends to $1$)

$$6x\sin\left (\frac{4}{x} \right )=6\frac{1}{\frac{1}{x}}\sin\left (\frac{4}{x} \right ) =24\frac{1}{\frac{4}{x}}\sin\left (\frac{4}{x} \right ) =24\frac{\sin\left (\frac{4}{x} \right )}{\frac{4}{x}}$$

Can you take it from there?

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