limit of $6x\sin(\frac{4}{x})$ as $x$ approaches infinity 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. 
 A: Make a change of variable: ($n = 1/x$)
$$\lim_{x\to \infty}6x\sin \frac 4x=\lim_{n\to 0}\frac{6\sin 4n}{n}$$
$$=\lim_{n\to 0}\frac{24\sin 4n}{4n}$$
$$=\lim_{n\to 0}24\left(\frac{\sin 4n}{4n}\right)$$
$$=24$$
A: 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
$$\lim_{x\to \infty}6x\sin\left(\frac{4}{x}\right)=24$$
A: $$\lim_{x\to\infty} 6x\sin\left(\frac{4}{x}\right) = \lim_{x\to\infty} 6x\left(\frac{4}{x}+O\left(x^{-3}\right)\right) = \lim_{x\to\infty} 24+O(x^{-2}) = 24$$
A: 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?
