$\lim_{\rightarrow \infty} \sin(x²)$
1 Answer
The following limit
\[\lim_{x\rightarrow \infty}\sin x^2\]
does not exists as $\sin (x^2)$ oscillates between $-1$ and $1$ as $x \rightarrow \infty.$ However,
\[0=\lim_{x\rightarrow \infty}-\frac{1}{x} \leq \lim_{x\rightarrow \infty}\frac{\sin x^2}{x} \leq \lim_{x\rightarrow \infty}\frac{1}{x}=0.\]
Hence
\[\lim_{x\rightarrow \infty}\frac{\sin x^2}{x}=0.\]
Savionf
574
-
thanks a lot
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 422 views
- Pro Bono
Related Questions
- Existence of a Divergent Subsequence to Infinity in Unbounded Sequences
- Hs level math (problem solving) *der
- Use Green’s theorem to compute $\int_C x^2 ydx − xy^2 dy$ where $C$ is the circle $x^2 + y ^2 = 4$ oriented counter-clockwise.
- Why does $ \sum\limits_{n=1}^{\infty } 2^{2n} \times \frac{(n!)^2}{n(2n+1)(2n)!} =2 $ ?
- Derivatives again. Thank you!
- Is $\int_0^{\infty}\frac{x+3}{x^2+\cos x}$ convergent?
- Show that the line integral $ \oint_C y z d x + x z d y + x y d z$ is zero along any closed contour C .
- Recursive square root sequence
