Tính lim sin n n 3 + 1
A. 0
B. + ∞
C. - ∞
D. 1
Những câu hỏi liên quan
Giúp e 3 bài này với ạ 1, Lim sin^2n / n + 2 2, Lim 1 + cosn / 2n + 3 3, Lim cosn + 4 / 5 + n
14 tháng 1 2022 lúc 17:21
Tính
lim \(\dfrac{sin\left(3n+n^2\right)}{3n-1}\)
\(-1\le sin\left(X\right)\le1\Rightarrow\dfrac{-1}{3n-1}\le f\le\dfrac{1}{3n-1}\)
Do \(\lim\left(-\dfrac{1}{3n-1}\right)=\lim\left(\dfrac{1}{3n-1}\right)=0\Rightarrow\lim\left(f\right)=0\) theo định lý kẹp
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3 tháng 2 2021 lúc 23:06
tìm \(lim\sqrt{\dfrac{3^n+2^{n+1}}{sin^n50^0+3^{n+2}}}\)
\(\lim\sqrt{\dfrac{3^n+2.2^n}{sin^n50^0+9.3^n}}=\lim\sqrt{\dfrac{1+2\left(\dfrac{2}{3}\right)^n}{\left(\dfrac{sin50^0}{3}\right)^n+9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
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Tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x\sin2x...\sin nx}{x^n}\).
Tính l i m n + sin n 3 n + 4 .
A. 0
B. 1 3
C. 1 4
D. + ∞
\(lim\frac{1}{n^3}.sin\frac{n^2\pi}{3}\)
Do \(-1\le sinx\le1\) \(\forall x\Rightarrow\frac{-1}{n^3}\le\frac{sin\frac{n^2\pi}{3}}{n^3}\le\frac{1}{n^3}\)
Mà \(lim\frac{-1}{n^3}=lim\frac{1}{n^3}=0\)
\(\Rightarrow lim\frac{1}{n^3}.sin\frac{n^2\pi}{3}=0\) theo giới hạn kẹp
tìm các giới hạn sau:
a; lim\(\frac{1+2+3+...+n}{3n^3}\)
b, lim \(\left(\frac{n+2}{n+1}+\frac{sin\text{n}}{2^n}\right)\)
c, lim \(\left(\sqrt{n^2-3n}-\sqrt{n^2+1}\right)\)
d,\(lim\left(\sqrt[3]{n^3+3n^2}-n\right)\)
\(a=lim\frac{n^2+n}{6n^3}=lim\frac{\frac{1}{n}+\frac{1}{n^3}}{6}=\frac{0}{6}=0\)
\(b=lim\frac{1+\frac{2}{n}}{1+\frac{1}{n}}+lim\frac{sinn}{2^n}=1+0=1\)
Giải thích: \(-1\le sin\left(n\right)\le1\) \(\forall n\Rightarrow\frac{-1}{2^n}\le\frac{sin\left(n\right)}{2^n}\le\frac{1}{2^n}\)
Mà \(lim\frac{-1}{2^n}=lim\frac{1}{2^n}=0\Rightarrow lim\frac{sin\left(n\right)}{2^n}=0\) theo nguyên tắc giới hạn kẹp
\(c=lim\frac{-3n-1}{\sqrt{n^2-3n}+\sqrt{n^2+1}}=lim\frac{-3-\frac{1}{n}}{\sqrt{1-\frac{3}{n}}+\sqrt{1+\frac{1}{n^2}}}=\frac{-3}{1+1}=-\frac{3}{2}\)
\(d=lim\frac{3n^2}{\sqrt[3]{\left(n^3+3n^2\right)^2}+n\sqrt[3]{n^3+3n^2}+n^2}=lim\frac{3}{\sqrt[3]{\left(1+\frac{3}{n}\right)^2}+\sqrt[3]{1+\frac{3}{n}}+1}=\frac{3}{1+1+1}=1\)
tìm giới hanjn
1) lim \(\frac{\left(-1\right)^n}{n-3}\)
2) lim \(\frac{n\left(sin\left(pi.n^2\right)\right)}{n^2+3n-2}\)
1) lim\(\frac{\left(-1\right)^n}{n-3}\)
ta có: \(\left|\frac{\left(-1\right)^n}{n-3}\right|=\frac{1}{n-3}< \frac{1}{n-4}\)
lim \(\frac{1}{n-4}=lim\frac{\frac{1}{n}}{1-\frac{4}{n}}=\frac{lim0}{1}=0\)
2) lim\(\frac{nsin\left(pi.n^2\right)}{n^2+3n-2}\)
ta có : \(\left|\frac{nsin\left(pi.n^2\right)}{n^2+3n-2}\right|\)<=\(\frac{n}{n^2+3n-2}\)
=> lim\(\frac{n}{n^2+3n-2}=0\)
=>lim\(\frac{nsin\left(pi.n^2\right)}{n^2+3n-2}\)=0
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3 tháng 3 2021 lúc 18:32
lim\(\dfrac{2\sqrt{n}-sin\sqrt{n}}{3\sqrt{n+2}+5}\)
Ta có: \(-\dfrac{1}{\sqrt{n}}\le\dfrac{sin\sqrt{n}}{\sqrt{n}}\le\dfrac{1}{\sqrt{n}}\) , mà
\(\lim\left(-\dfrac{1}{\sqrt{n}}\right)=\lim\left(\dfrac{1}{\sqrt{n}}\right)=0\Rightarrow\lim\left(\dfrac{sin\sqrt{n}}{\sqrt{n}}\right)=0\)
Do đó:
\(\lim\dfrac{2\sqrt{n}-sin\sqrt{n}}{3\sqrt{n+2}+5}=\lim\dfrac{2-\dfrac{sin\sqrt{n}}{\sqrt{n}}}{3\sqrt{1+\dfrac{2}{n}}+\dfrac{5}{\sqrt{n}}}=\dfrac{2-0}{3+0}=\dfrac{2}{3}\)
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Tính các giới hạn
a) \(\lim\limits_{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a}\)
b) \(\lim\limits_{x\rightarrow1}\left(1-x\right)\tan\dfrac{\pi x}{2}\)
c) \(\lim\limits_{x\rightarrow\dfrac{\pi}{3}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\)
d) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
