Question

[Phụng Nguyễn Thị]

Tính các giới hạn sau đây :

\(L_1=lim\frac{x^3+3x^2-2x}{x^5+4x}\left(x\rightarrow0\right)\)

\(L_2=lim\frac{x^3-3x+2}{\left(4-2x\right)^3}\left(x\rightarrow+\infty\right)\)

\(L_3=lim\frac{2x^2+3x+1}{x^2+x}\left(x\rightarrow-1\right)\)

\(L_4=lim\frac{x^2-4x+1}{4-x^2}\left(x\rightarrow2\right)\)

\(L_5=lim\frac{\sqrt{x+1}-2}{x-2}\left(x\rightarrow3\right)\)

\(L_6=lim\frac{\sqrt{x+3}-x-1}{x^2-1}\left(x\rightarrow1\right)\)

\(L_7=lim\left(\sqrt{x^2+x+1}-x+1\right)\left(x\rightarrow+\infty\right)\)

\(L_8=lim\left(\sqrt{x^2+x+1}-3x+2\right)\left(x\rightarrow-\infty\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(L_1=\lim\limits_{x\rightarrow0}\frac{x\left(x^2+3x-2\right)}{x\left(x^4+4\right)}=\lim\limits_{x\rightarrow0}\frac{x^2+3x-2}{x^4+4}=-\frac{1}{2}\)

\(L_2=\lim\limits_{x\rightarrow+\infty}\frac{1-\frac{3}{x^2}+\frac{2}{x^3}}{\left(\frac{4}{x}-2\right)^3}=\frac{1}{\left(-2\right)^3}=-\frac{1}{8}\)

\(L_3=\lim\limits_{x\rightarrow-1}\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{2x+1}{x}=1\)

\(L_4=\lim\limits_{x\rightarrow2}\frac{x^2-4x+1}{4-x^2}=\frac{1}{0}=+\infty\)

\(L_5=\lim\limits_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-2}=\frac{0}{1}=0\)

\(L_6=\lim\limits_{x\rightarrow1}\frac{x+3-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{-\left(x+2\right)}{\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\frac{-3}{2.4}=-\frac{3}{8}\)

\(L_7=\lim\limits_{x\rightarrow+\infty}\frac{x^2+x+1-\left(x-1\right)^2}{\sqrt{x^2+x+1}+x-1}\lim\limits_{x\rightarrow+\infty}\frac{3x}{\sqrt{x^2+x+1}+x-1}=\lim\limits_{x\rightarrow+\infty}\frac{3}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1-\frac{1}{x}}=\frac{3}{2}\)

\(L_8=\lim\limits_{x\rightarrow-\infty}\frac{x^2+x+1-\left(3x-2\right)^2}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8x^2+13x-3}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8+\frac{13}{x}-\frac{3}{x^2}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+3-\frac{2}{x}}=\frac{-8}{-1+3}=-4\)