Question

Tìm \(\lim\limits_{x->-\infty}\)\(\frac{\left|x\right|\sqrt{4x^2+3}}{2x-1}\)

lim \(\sqrt{n}\)(\(\sqrt{n+4}\)-\(\sqrt{n+3}\))

lim (n-2-\(\sqrt{3n^2+n-1}\))

\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)

Answer 1

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

\(lim\frac{\sqrt{n}}{\sqrt{n+4}+\sqrt{n+3}}=lim\frac{1}{\sqrt{1+\frac{4}{n}}+\sqrt{1+\frac{3}{n}}}=\frac{1}{2}\)

\(lim\left(\frac{\left(n-2\right)^2-\left(3n^2+n-1\right)}{n-2+\sqrt{3n^2+n-1}}\right)=lim\frac{-2n^2-5n+5}{n-2+\sqrt{3n^2+n-1}}=lim\frac{-2n+5+\frac{5}{n}}{1-\frac{2}{n}+\sqrt{3+\frac{1}{n}-\frac{1}{n^2}}}=-\infty\)

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