I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} – 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
Best Answer
Here's an alternative answer if you absolutely have to use L'Hopital's rule:
First rewrite the expression inside the limit as follows:
$$x^3\Big(\sin(\frac{1}{x+2})-2\sin(\frac{1}{x+1})+\sin(\frac{1}{x})\Big)=x^3\Big[(\sin(\frac{1}{x+2})-\frac{1}{x+2})-2(\sin(\frac{1}{x+1})-\frac{1}{x+1})+\sin(\frac{1}{x})\Big]+x^3\Big(\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}\Big)$$
We've written the expression in this way, suggestively, so that for each individual term in parentheses the limits exist. Then we compute the limits as follows:
$$\lim_{x\to\infty}x^3(\sin(\frac{1}{x})-\frac{1}{x})=\lim_{u\to 0}\frac{\sin(u)-u}{u^3}=-\frac{1}{6}$$
by applying L'Hopital's rule twice.
Also
$$\lim_{x\to\infty}x^3(\sin(\frac{1}{x+1})-\frac{1}{x+1})=\Big[\lim_{x\to\infty}(\frac{x}{x+1})^3\Big]\Big[\lim_{x\to\infty}(x+1)^3(\sin(\frac{1}{x+1})-\frac{1}{x+1})\Big]=-\frac{1}{6}$$ and similarly $$\lim_{x\to\infty}x^3(\sin(\frac{1}{x+2})-\frac{1}{x+2})=-\frac{1}{6}$$
Finally
$$\lim_{x\to\infty}x^3\Big(\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}\Big)=\lim_{x\to\infty}\frac{2x^3}{x(x+1)(x+2)}=2$$
and hence we find by adding all those limits together that
$$\lim_{x\to\infty}x^3\Big(\sin(\frac{1}{x+2})-2\sin(\frac{1}{x+1})+\sin(\frac{1}{x})\Big)=-\frac{1}{6}+2\frac{1}{6}-\frac{1}{6}+2=2$$
The takeaway from this manipulation is that applying L'Hopital's rule is not straightforward, but there is a way avoid lengthy calculations , by which one has to add and subtract terms that amount to known or easily derived limits. However, in my personal opinion, expanding in a Taylor series is the only foolproof prescription for taking limits of that sort.