\(lim_{x\rightarrow0}\frac{tan2x-sin2x}{x^3}\)
Giới hạn nào sau đây tồn tại:
A, \(lim_{x\rightarrow+\infty}sin2x\) B, \(lim_{x\rightarrow+\infty}cos3x\) C, \(lim_{x\rightarrow0}sin\frac{1}{2x}\) D, \(lim_{x\rightarrow1}sin\frac{1}{2x}\)
Đáp án D đúng
Tìm các giới hạn sau:
a) \(lim_{x\rightarrow0}\dfrac{tan3x}{sin5x}\)
b) \(lim_{x\rightarrow0}\dfrac{cos2x-1}{sin^23x}\)
c) \(lim_{x\rightarrow1}\dfrac{x^2-4x+3}{sin\left(x-1\right)}\)
\(lim_{x\rightarrow0}xcos\frac{1}{x}\)
\(\left|cos\frac{1}{x}\right|\le1\Rightarrow\left|x.cos\frac{1}{x}\right|\le\left|x\right|\)
Mà \(\lim\limits_{x\rightarrow0}\left|x\right|=0\Rightarrow\lim\limits_{x\rightarrow0}x.cos\frac{1}{x}=0\)
\(lim_{x\rightarrow0}\frac{\left(1+3x\right)^3+\left(1-4x\right)^4}{x}\)
\(\sqrt{3}\sin^2\left(x\right)+\frac{1}{2}\sin2x=\tan2x\)\(\text{}\sqrt{3}\sin^2\left(x\right)+\frac{1}{2}\sin2x=\tan2x\)
\(D=lim_{x\rightarrow0}\frac{\left(1+2x\right)^2\left(1+3x\right)^3-1}{x}\)
Xác định \(lim_{x\rightarrow0}\frac{\left|x\right|}{x^2}\)
\(lim_{x\rightarrow0}\dfrac{\sqrt{x^3+1}-1}{x^2+x}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^3+1}-1}{x^2+x}\\ =\lim\limits_{x\rightarrow0}\dfrac{x\sqrt{x+\dfrac{1}{x^2}}-1}{x\left(x+1\right)}\\ =\lim\limits_{x\rightarrow0}\dfrac{x\left(\sqrt{x+\dfrac{1}{x^2}}-\dfrac{1}{x}\right)}{x\left(x+1\right)}\\ =\dfrac{\sqrt{x+\dfrac{1}{x^2}}-\dfrac{1}{x}}{x+1}\\ =\dfrac{\sqrt{x}}{x+1}=\dfrac{0}{0+1}=0\)
\(\lim_{x\rightarrow0}\frac{\left|x\right|}{x}\)
giúp vs nhé