2021年10月20日 · \begin {align} \lim_ {x \to \infty} n\sin (2\pi en!)&=\lim_ {x \to \infty}n\sin [2\pi n! (1+\frac12+...+\frac1 {n!}+\frac1 { (n+1)!}+\frac {e^\theta} { (n+2)!})] (e^\theta\in (1,e))\\ &=\lim_ …

Assume $\lim \sin(n) = l$. Then so is $\lim \sin(2n) = l$. So $\lim \cos(2n) = 1 - 2l^2$, but so does the limit of $\cos(2(n + 1))$. Now apply the sum-formula to $\sin(2(n + 1) - 2n)$.

For the sine function in degrees, the answer is that the limit is zero. I can say this because for every n ≥ 360 n ≥ 360, 360 360 divides n! n!. And if 360 360 divides the number, then the sine of that number is zero. For the sine function that uses radians, I can't think how to prove it at the moment, but I suspect the function does not converge.

2021年2月7日 · 当n趋近于无穷时，sinn极限不存在。 你认为这个结论正确的前提只有一个：当n趋近于无穷。 因此，当你发现nπ也趋近于无穷，也满足此前提，于是直接套用上述结论，得出sinnπ的极限不存在的结论。

很简单的反例就是An＝sin (2nπ+1/n)，虽然2nπ+1/n趋于无穷，但显然An的极限是为0的。

2024年11月29日 · Simply put if $\lim\limits_{n→∞} f(n)$ exists, then $\lim\limits_{n→∞} f '(n) = 0$. In the given function, the derivative varies periodically so the limit does not exist.

输入识别各种同义词像功能 asin, arsin, arcsin, sin^-1. 乘号和括号被额外放置 - 写 2sinx 相似的 2*sin(x) 数学函数和常数列表: • ln(x) — 自然对数. • sin(x) — 正弦. • cos(x) — 余弦. • tan(x) — …

Free Limit Calculator helps you solve one-dimensional and multivariate limits for calculus and mathematical analysis. Get series expansions and graphs.

The perimeter of the polygon is $n*s$, where $n$ is the number of sides the polygon has, and $s$ is the side length of the polygon. $s$ is related to $r$, the circles radius, by $\frac{s}{2} = r*\sin(\frac{2\pi}{2n})$.

原式 =\\lim\\limits_{n\\rightarrow\\infty}\\sin^2(\\pi \\sqrt{n^2+n}-n\\pi) =\\lim\\limits_{n\\rightarrow\\infty}\\sin^2[n\\pi (\\sqrt{1+\\frac1n}-1)] =\\lim\\limits_{n\\rightarrow\\infty}\\sin^2\\frac{\\pi}{2} =1