# 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.

• How about $\sin x \approx x$ for very small $x$ ?
– LHF
Commented Feb 21, 2020 at 20:12

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.

• +1 for smart approach. Commented Feb 22, 2020 at 1:44

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,

$$\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t} + \sin t \right )=\lim_{t \to 0} \frac{2t^3+O(t^4)} {t^3}=2$$

• I almost understood all of your solution. If you don't mind, could you explain how did you get from $\sin(t-t^2+2t^3) \cos t^2$ to $\sin(t-t^2+2t^3)$ (from your $4$th line to your $5$th line, counting the lines where you wrote expressions). What happened to the $\cos t^2$? The second confusion I got was on your $6$th line. What happened there? How did you get from $4\cos t \sin \dfrac{t^3}{2}$ to $2t^3$? If you could clear up these points, I would really appreciate it.
– user592938
Commented Feb 22, 2020 at 22:24

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} \

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...

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}