\(\lim\limits_{x\rightarrow0}\dfrac{cos\sqrt{x}-1}{ln\left(x+1\right)}=-\dfrac{1}{2}\).

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 \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

thầy cô và các bạn biết câu nào giúp mình câu đó em rất cảm ơn ạ

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}-\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}+\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{3x}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{2}{x^2}}=0\)

\(-1\le\sin2x\le1\Rightarrow\dfrac{-5}{x^2+2}\le\dfrac{5\sin2x}{x^2+2}\le\dfrac{5}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}-\dfrac{5}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{5}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}=0\)

\(0\le\cos^2x\le1\Rightarrow0\le\dfrac{\cos^2x}{x^2+2}\le\dfrac{1}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}=0\)

\(\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{3x-5\sin2x+\cos^2x}{x^2+2}=0\)

Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :

\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)

    \(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)