Lời giải:
\(\lim_{x\to +\infty}\frac{\sqrt{x^2+1}}{x}=\lim_{x\to +\infty}\sqrt{\frac{x^2+1}{x^2}}=\lim_{x\to +\infty}\sqrt{1+\frac{1}{x^2}}\)
\(=\sqrt{1}=1\)
\(\lim\limits_{x\rightarrow0}\left(\frac{1}{x}+\frac{1}{x^2}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{\left(-x\right)^2+2}}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\frac{|x|+\sqrt{x^2+x}}{x+10}\)
\\(\\lim\\limits_{x\\rightarrow8}\\frac{\\sqrt[3]{x}-2}{2x-16}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow-2}\\frac{\\sqrt{x-3}-1}{\\sqrt[3]{x-6}+2}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow1}\\frac{2x-1-\\sqrt{x^2+2x-2}}{x^2-4x+3}\\)
\n\(\lim\limits_{x\rightarrow4}\frac{2x-\sqrt{3x+1}}{x^2-1}\)
\(\lim\limits_{x\rightarrow8}\frac{\sqrt[3]{x}-\sqrt{x-4}}{x-8}\)
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-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)
tính
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}\)
Tính: \(\lim\limits_{x\rightarrow1}\frac{\sqrt{3x-2}+\sqrt[3]{3x+5}-\frac{7}{4}x-\frac{5}{4}}{x^2-2x+1}\)
\(\lim\limits_{x\rightarrow0}\frac{\sqrt{2x+1}-\sqrt[3]{3x+1}}{x^2}\)
\(\lim\limits_{x\rightarrow0}\frac{\sqrt[2019]{x^3+x^2+x+1}-1}{x}\)
