Bạn đang xem: What is the limit of sinx as x approaches infinity?
I understand -1
Also, if you use the L"hopital rule instead of squeeze theorem for sin(2x)/x you get it is equal to limit of 2sin(2x)/1. 2sin(2x)/1 as x goes to infinity is undefind ! So squeeze theorem says the original limit is 0 while the L Hoptial rule says the original limit is undefined. Which rule do you use?
Thank you so much.
The easiest use of the squeeze theorem for $\lim_{x\to\pm\infty}\frac{\sin f(x)}{x}$ is $-\frac{1}{|x|}\le\frac{\sin f(x)}{x}\le\frac{1}{|x|}$, so the limit is $0$.
If you are taking $x \to \infty$ you don"t have to worry about the case where $x$ is negative.
You cannot apply l"Hopital"s rule because the numerator $\sin(2x)$ does not have a limit as $x \to \infty$.
Thanks for contributing an answer to nasaconstellation.comematics Stack Exchange!
Please be sure to answer the question. Provide details and share your research!But avoid …
Asking for help, clarification, or responding to other answers.Making statements based on opinion; back them up with references or personal experience.Use nasaconstellation.comJax to format equations. nasaconstellation.comJax reference.
Xem thêm: Nghĩa Của Từ Review Nghĩa Là Gì ? Cách Để Viết Bài Review Xuất Sắc Nhất
To learn more, see our tips on writing great answers.
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy
Not the answer you're looking for? Browse other questions tagged limits or ask your own question.
Limit of $\sin\left(\frac{\pi}{x} \right)$ as $x$ approaches $0$ does not exist, with squeeze theorem
Site design / logo © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. rev2022.4.14.41981
Your privacy
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.