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

\(=\lim\limits_{x\rightarrow\infty}2cos\left(\dfrac{\sqrt{x+1}+\sqrt{x}}{2}\right)sin\left(\dfrac{\sqrt{x+1}-\sqrt{x}}{2}\right)\)

\(=\lim\limits_{x\rightarrow\infty}2cos\left(\dfrac{\sqrt{x+1}+\sqrt{x}}{2}\right)sin\left(\dfrac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}\right)\)

Ta có:

\(-2\le2cos\left(\dfrac{\sqrt{x+1}+\sqrt{x}}{2}\right)\le2\) (hữu hạn)

\(\lim\limits_{x\rightarrow\infty}sin\left(\dfrac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}\right)=sin\left(0\right)=0\)

\(\Rightarrow\lim\limits_{x\rightarrow\infty}\left(sin\sqrt{x+1}-sin\sqrt{x}\right)=0\)

\(\left\{{}\begin{matrix}-1\le sinx\le1\\-3\le3cos2x\le3\end{matrix}\right.\) \(\Rightarrow-4\le sinx+3cos3x\le4\) (dấu = có xảy ra hay ko ko hề quan trọng)

\(\Rightarrow\frac{-4}{x^2-2x+3}\le\frac{sinx+3cos2x}{x^2-2x+3}\le\frac{4}{x^2-2x+3}\)

Mà \(\lim\limits_{x\rightarrow\infty}\frac{-4}{x^2-2x+3}=\lim\limits_{x\rightarrow\infty}\frac{4}{x^2-2x+3}=0\)

\(\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx+3cos2x}{x^2-2x+3}=0\)

Oke cám mơn nha

a/ \(\Leftrightarrow\left[{}\begin{matrix}a>1\\\frac{a+1}{2}< -1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a>1\\a< -3\end{matrix}\right.\)

b/ \(\left(-\infty;5\right)\cup\left(-3;+\infty\right)=R\) nên với mọi a thì \(\left[a;\frac{a+1}{2}\right]\in\left(-\infty;5\right)\cup\left(-3;+\infty\right)\)

\(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)

\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)=\left(-2;7\right)\)

\(\left(0;12\right)\cap[5;+\infty)=[5;12)\)

\(R\cap\left[-1;1\right]=\left[-1;1\right]\)