Bài 2: Giới hạn của hàm số

\(\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{1-x}+x-1}{\sqrt{x^2-x^3}}\)

\(=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{1-x}-\left(1-x\right)}{x\cdot\sqrt{1-x}}\)

\(=\lim\limits_{x\rightarrow1^-}\dfrac{1-\sqrt{1-x}}{x}\)

\(=\dfrac{1-\sqrt{1-1}}{1}=\dfrac{1}{1}=1\)

\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x^3}{3x^2-4}-\dfrac{x^2}{3x+2}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^3\left(3x+2\right)-x^2\left(3x^2-4\right)}{\left(3x^2-4\right)\left(3x+2\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{3x^4+2x^3-3x^4+4x^2}{\left(3x^2-4\right)\left(3x+2\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x^3+4x^2}{\left(3x^2-4\right)\left(3x+2\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2+\dfrac{4}{x}}{\left(3-\dfrac{4}{x^2}\right)\left(3+\dfrac{2}{x}\right)}\)

\(=\dfrac{2+0}{\left(3-0\right)\left(3+0\right)}=\dfrac{2}{9}\)

Th1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)

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

\(=\dfrac{1}{\sqrt{1+0+0}}=\dfrac{1}{1}=1\)

TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x}{-x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}=\dfrac{-1}{1}=-1\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\cdot\sqrt{1+\dfrac{1}{x^2}}}{x+1-x\cdot\sqrt{1+\dfrac{1}{x^2}}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}-\sqrt{1+\dfrac{1}{x^2}}}\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}-\sqrt{1+\dfrac{1}{x^2}}=-1< 0\\\lim\limits_{x\rightarrow-\infty}1+\dfrac{1}{x}-\sqrt{1+\dfrac{1}{x^2}}=1+0-\sqrt{1+0}=0\end{matrix}\right.\)

TH1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x+1}{\sqrt{x^2-x+1}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{x}}{\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}}\)

\(=\dfrac{1+0}{\sqrt{1-0+0}}=\dfrac{1}{1}=1\)

TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+1}{\sqrt{x^2-x+1}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x+1}{-x\cdot\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}}\)

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

TH1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{2+\dfrac{3}{x^2}}}{4+\dfrac{2}{x}}\)

\(=\dfrac{\sqrt{2+0}}{4+0}=\dfrac{\sqrt{2}}{4}\)

TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{2+\dfrac{3}{x^2}}}{4+\dfrac{2}{x}}\)

\(=\dfrac{-\sqrt{2+0}}{4+0}=-\dfrac{\sqrt{2}}{4}\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{3x+2\sqrt{x}-1}{x-2\sqrt{x}+4}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{3+\dfrac{2}{\sqrt{x}}-\dfrac{1}{x}}{1-\dfrac{2}{\sqrt{x}}+\dfrac{4}{x}}\)

\(=\dfrac{3+0-0}{1-0+0}=\dfrac{3}{1}=3\)

TH1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x}{1+\left|x\right|}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{1+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\dfrac{1}{x}+1}=\dfrac{1}{0+1}=1\)

TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x}{1+\left|x\right|}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x}{1-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1}{\dfrac{1}{x}-1}=\dfrac{1}{0-1}=-1\)

Đề sai.

\(\lim\limits_{x\rightarrow0}\dfrac{2x}{\sqrt{4x^2+x^3}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2x}{x\cdot\sqrt{4+x}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{4+x}}=\dfrac{2}{\sqrt{4+0}}=1\)

Không tồn tại do \(\lim\limits_{x\rightarrow0^-}\dfrac{2x}{\sqrt{4x^2+x^3}}\ne\lim\limits_{x\rightarrow0^+}\dfrac{2x}{\sqrt{4x^2+x^3}}\)