# I bogged down with $~\lim_{x\to\infty}\frac{x^{4}}{2}\left(\frac{1}{x^{2}}-\sin^{}\left(\frac{1}{x^{2}}\right)\right)~$

$$A:=\lim_{x\to\infty}\frac{x^{4}}{2}\left(\frac{1}{x^{2}}-\sin^{}\left(\frac{1}{x^{2}}\right)\right)\tag{1}$$

$$=\frac{1}{2}\lim_{x\to\infty}x^{4}\left(\frac{1}{x^{2}}-\sin^{}\left(\frac{1}{x^{2}}\right)\right)$$

$$=\frac{1}{2}\lim_{x\to\infty}\left(x^{2}-x^{4}\sin^{}\left(\frac{1}{x^{2}}\right)\right)$$

$$=\frac{1}{2}\left\{\lim_{x\to\infty}x^{2}-\lim_{x\to\infty}x^{4}\sin^{}\left(\frac{1}{x^{2}}\right)\right\}$$

So, obiviously the left limit of above diverges to infinity so the attracting part is the remaining right limit.

$$B:=\lim_{x\to\infty}x^{4}\sin^{}\left(\frac{1}{x^{2}}\right)$$

$$=\lim_{x\to\infty}x^{4}\sin^{}\left(x^{-2}\right)$$

$$=\lim_{x\to\infty}x^{4}\left(\sin^{}\left(x^{-2}\right)^{-1}\right)^{-1}$$

$$=\lim_{x\to\infty}x^{4}\left(\csc\left(x^{-2}\right)\right)^{-1}$$

$$=\lim_{x\to\infty}\frac{x^{4}}{\csc\left(x^{-2}\right)}$$

$$=\lim_{x\to\infty}\frac{\frac{x^{4}}{1}}{\frac{1}{\sin^{}\left(x^{-2}\right)}}$$

$$=\lim_{x\to\infty}\frac{x^{4}/1}{\frac{1}{\sin^{}\left(\frac{1}{x^{2}}\right)}}$$

The nominator$$~x^{4}/1~$$diverges to infinity, and,$$~\sin^{}\left(\frac{1}{x^{2}}\right)~$$converges to 0, which means the reciprocal$$~\left(1/\sin^{}\left(\frac{1}{x^2}\right)\right)~$$of it diverges to infinity.

Hence the theorem of L'hopital can be used here.

$$B=\lim_{x\to\infty}\frac{4x^{3}}{\frac{d}{dx}\left(\sin^{}\left(x^{-2}\right)^{-1}\right)}$$

$$\frac{d}{dx}\left(\sin^{}\left(x^{-2}\right)^{-1}\right)$$

$$=\left(-1\right)\sin^{}\left(x^{-2}\right)^{-2}\left\{\left(-2\right)x^{-3}\right\}$$

$$=2\cdot\sin^{}\left(x^{-2}\right)^{-2}x^{-3}$$

$$=\frac{2}{x^{3}\sin^{}\left(x^{-2}\right)^{2}}$$

$$=\frac{2}{x^{3}\sin^{}\left(\frac{1}{x^{2}}\right)^{2}}$$

About the denominator above,$$~x^{3}~$$diverges to infinity and$$~\sin^{}\left(\frac{1}{x^{2}}\right)^{2}~$$converges to zero,so seemingly no such any progress is gained so far by my these approaches.

I think I have to do another approach.

I will watch this first since I have no experience of big O notation with limit of math so far.

• If you don't want to use Taylo polynomial, notice that you can apply L'hopital rules twice in the answer of @Parcly Taxel...
– Surb
Nov 6, 2021 at 13:34

Set $$y=\frac1{x^2}$$ to get $$A=\lim_{y\to0}\frac1{2y^2}(y-\sin y)$$ Now note that $$y-\sin y\in O(y^3)$$, so the expression inside the limit is $$O(y)$$ and $$A=0$$.
• Note that $$lim_{x\to a}\left(f(x)+g(x)\right)=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$$ iff, $lim_{x\to a}f(x), \lim_{x\to a}g(x)$ are converging. I think you violated that. Nov 6, 2021 at 12:39
• @ParclyTaxel I solved the limit problem $~A~$ using L'hopital theorem, without Landau big O notation. I think transformation of variable of lim is least required to solve this limit. Still , I can't understand why $~x-\sin(x)\in O(y^3) ~$can be held though.. Nov 28, 2021 at 5:58