Tính: \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
Những câu hỏi liên quan
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) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
Tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x-\tan x}{\left(\sqrt{1+2x}-\sqrt[3]{1+3x}\right)\cdot\sin10x}\)
Xem chi tiết
tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x-\sqrt{3}\cos5x}{3x}\)
Tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x\sin2x...\sin nx}{x^n}\).
\(\lim\limits_{x\rightarrow0}\dfrac{\sin x-\sqrt{3}\cos5x}{3x}\)
\(=\dfrac{1}{3}\lim\limits_{x\rightarrow0}\dfrac{sinx}{x}-\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}=\dfrac{1}{3}-\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}\)
Xét:
\(\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{3}cos5x}{3x}=\dfrac{\sqrt{3}}{0}=+\infty\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{-\sqrt{3}cos5x}{-3x}=\dfrac{-\sqrt{3}}{0}=-\infty\)
\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}\) ko tồn tại nên giới hạn đã cho không tồn tại
Đúng 1
Bình luận (0)
Tính \(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sin x}-\dfrac{3}{\sin3x}\right)\dfrac{1}{x}\)
tính \(\lim\limits_{x\rightarrow0}\left(\dfrac{1-\sin x-\cos x}{1+\sin3x+\cos3x}\right)\)
\(I=\lim\limits_{x\rightarrow0}\dfrac{^{ }e^{\sin^2\left(ax\right)}\left(1-\sqrt{1+x^2}\right)}{x^2\tan\left(bx\right)}\)
Bài 1
a. limlimits_{xrightarrow+infty}frac{1+2sqrt{x}-x}{x+3} b. limlimits_{xrightarrow+infty}frac{x^3+3x-1}{x^2sqrt{x}+x} c. limlimits_{xrightarrow-infty}frac{x+2sqrt{1-x}}{1-x}
Bài 2: Tính các giới hạn sau biết limlimits_{xrightarrow0}frac{sin x}{x}1
a. limlimits_{xrightarrow0}frac{1-cos x}{1-cos3x} b. limlimits_{xrightarrow0}frac{cot x-sin x}{x^3} c. limlimits_{xrightarrowinfty}frac{x.sin x}{2x^2}
Đọc tiếp
Bài 1
a. \(\lim\limits_{x\rightarrow+\infty}\frac{1+2\sqrt{x}-x}{x+3}\) b. \(\lim\limits_{x\rightarrow+\infty}\frac{x^3+3x-1}{x^2\sqrt{x}+x}\) c. \(\lim\limits_{x\rightarrow-\infty}\frac{x+2\sqrt{1-x}}{1-x}\)
Bài 2: Tính các giới hạn sau biết \(\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1\)
a. \(\lim\limits_{x\rightarrow0}\frac{1-\cos x}{1-\cos3x}\) b. \(\lim\limits_{x\rightarrow0}\frac{\cot x-\sin x}{x^3}\) c. \(\lim\limits_{x\rightarrow\infty}\frac{x.\sin x}{2x^2}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)
Bài 2:
\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)
\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)
Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)
Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)