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








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.







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 need your wisdom.


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

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

1 Answer 1


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 $A=0$.

  • $\begingroup$ Is that big O notation same with time complexity of computer science? $\endgroup$ Nov 6, 2021 at 12:04
  • 1
    $\begingroup$ @electricalapprentice Yes. I am also a computer science undergrad (as of this post). $\endgroup$ Nov 6, 2021 at 12:05
  • $\begingroup$ 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. $\endgroup$
    – RAHUL
    Nov 6, 2021 at 12:39
  • $\begingroup$ @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.. $\endgroup$ Nov 28, 2021 at 5:58

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