Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ 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.
 A: Let $t=\frac1x$. Then,
$$\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
=\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t \right )$$
Use $\frac 1{1+a} = 1-a+a^2+O(a^3)$ to expand,
$$\sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t$$
$$=\sin(t-2t^2+4t^3)+\sin t  - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2\sin(t-t^2+2t^3)\cos t^2 - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2[\sin(t-t^2+2t^3) - \sin(t-t^2+t^3)]+O(t^4)$$
$$=4\cos t\sin\frac{t^3}2+O(t^4)= 4\cdot 1\cdot \frac{t^3}2+O(t^4)=2t^2+O(t^4)$$
where $\cos t^2 = 1 + O(t^4)$ is applied. Thus,
$$\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t} + \sin t \right )=\lim_{t \to 0}  \frac{2t^3+O(t^4)} {t^3}=2$$
A: 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.
A: You can't, because you need to work with the interplay of the sine functions. Concretely, using Taylor approximations (and collecting the errors together), 
\begin{align}
\sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} 
&=\frac1{x+2}-\frac1{6(x+2)^3}-2\left(\frac1{x+1}-\frac1{6(x+1)^3} \right)\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ +\frac1x-\frac1{6x^3}+o(\frac1{x^5})\\ \ \\
&=\frac2{x(x+1)(x+2)}-\frac1{6(x+2)^3}+\frac2{6(x+1)^3}\\ \ \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \  -\frac1{6x^3}+o(\frac1{x^5}).
\end{align}
Then
\begin{align}
x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
&=\frac2{(1+\tfrac2x)(1+\tfrac2x)}-\frac1{6(1+\tfrac2x)^3}\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac1{3(1+\tfrac1x)^3)}-\frac1{6x^3}+o(\tfrac1{x^2})\\ \ \\
&\xrightarrow[\vphantom{x_A}x\to\infty]{}2-\tfrac16+\tfrac13-\tfrac16=2.
\end{align}
\ 
A: Let $y=x+1$ then
$$\begin{align}\sum\sin&=\sin\left(\frac1{x+2}\right)-2\sin\left(\frac1{x+1}\right)+\sin\left(\frac1x\right)\\
&=\sin\left(\frac1y-\frac1{y^2}+\frac1{y^2(y+1)}\right)-2\sin\left(\frac1y\right)+\sin\left(\frac1y+\frac1{y^2}+\frac1{y^2(y-1)}\right)\\
&=\sin\left(\frac1{y^2(y+1)}\right)\cos\left(\frac1y-\frac1{y^2}\right)+\left(1-2\sin^2\left(\frac1{2y^2(y+1)}\right)\right)\\
&\quad\times\left(\sin\left(\frac1y\right)\left(1-2\sin^2\left(\frac1{2y^2}\right)\right)-\cos\left(\frac1y\right)\sin\left(\frac1{y^2}\right)\right)-2\sin\left(\frac1y\right)\\
&\quad+\sin\left(\frac1{y^2(y-1)}\right)\cos\left(\frac1y+\frac1{y^2}\right)+\left(1-2\sin^2\left(\frac1{2y^2(y-1)}\right)\right)\\
&\quad\times\left(\sin\left(\frac1y\right)\left(1-2\sin^2\left(\frac1{2y^2}\right)\right)+\cos\left(\frac1y\right)\sin\left(\frac1{y^2}\right)\right)\\
&=\sin\left(\frac1{y^2(y+1)}\right)\cos\left(\frac1y-\frac1{y^2}\right)-4\sin\left(\frac1y\right)\sin^2\left(\frac1{2y^2}\right)\\
&\quad-2\sin^2\left(\frac1{2y^2(y+1)}\right)\sin\left(\frac1y-\frac1{y^2}\right)-2\sin^2\left(\frac1{2y^2(y-1)}\right)\sin\left(\frac1y+\frac1{y^2}\right)\\
&\quad+\sin\left(\frac1{y^2(y-1)}\right)\cos\left(\frac1y+\frac1{y^2}\right)\end{align}$$
So
$$\begin{align}\lim_{x\rightarrow\infty}x^3\sum\sin&=\lim_{y\rightarrow\infty}\left\{\left(1+\frac1y\right)^2\frac{\sin\left(\frac1{y^2(y+1)}\right)}{\frac1{y^2(y+1)}}\cos\left(\frac1y-\frac1{y^2}\right)\right.\\
&\quad-\frac1{y^2}\left(1+\frac1y\right)^3\frac{\sin\left(\frac1y\right)}{\frac1y}\frac{\sin^2\left(\frac1{2y^2}\right)}{\left(\frac1{2y^2}\right)^2}\\
&\quad-\frac{1-\frac1{y^2}}{2y^4}\frac{\sin^2\left(\frac1{2y^2(y+1)}\right)}{\left(\frac1{2y^2(y+1)}\right)^2}\frac{\sin\left(\frac1y-\frac1{y^2}\right)}{\frac1y-\frac1{y^2}}\\
&\quad-\frac{\left(1+\frac1y\right)^4}{2y^4\left(1-\frac1y\right)^2}\frac{\sin^2\left(\frac1{2y^2(y-1)}\right)}{\left(\frac1{2y^2(y-1)}\right)^2}\frac{\sin\left(\frac1y+\frac1{y^2}\right)}{\frac1y+\frac1{y^2}}\\
&\quad\left.+\left(1+\frac1y\right)^2\frac{\sin\left(\frac1{y^2(y-1)}\right)}{\frac1{y^2(y+1)}}\cos\left(\frac1y+\frac1{y^2}\right)\right\}\\
&=1-0-0-0+1=2\end{align}$$
I just wanted to see how this looked in brute force trigonometric identities...
A: Using first approximation for $\sin x\approx x$ for $x$ near $0$, the limit can be rewritten without change of variables as $$\begin{aligned} &\lim_{x\to \infty}x^3\left(\frac{1}{x+2}-\frac{2}{x+1}+\frac{1}{x}\right)\\ = &\lim_{x\to \infty}x^3\left[\left(\frac{1}{x}-\frac{1}{x+1}\right)-\left(\frac{1}{x+1}-\frac{1}{x+2}\right)\right]\\=&\lim_{x\to \infty}x^3\left[\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}\right]\\=&\lim_{x\to \infty}\frac{2x^3}{x(x+1)(x+2)}\to 2\end{aligned}$$
