\(lim_{x->\pm\infty}\sqrt{x^2-3x+4}\)
\(lim_{x->\pm\infty}x\left(\sqrt{x^2+5}+x\right)\)
\(lim_{x->2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(lim_{x->0}\frac{x.sin2x}{1-cos2x}\)
\(lim_{x->0}\frac{\sqrt{1-x}-1}{x}\)
\(lim_{x->0-}\frac{1}{x}\left(\frac{1}{x+1}-1\right)\)
\(lim_{x->0-}\frac{2x+\sqrt{-x}}{5x-\sqrt{-x}}\)
\(lim_{x->1}\frac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{\left(x-1\right)^2}\)
l\(lim_{x->0}\left(1-x\right)tan\frac{\pi x}{2}\)
\(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-80}{x-3}=5\). Tính \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt[4]{f\left(x\right)+1}-3}{2x^2-11x+15}\)
\(lim_{x\rightarrow2}\frac{\left(\sqrt{x^2+6}-2x\right)\left(\sqrt{4x+1}+x\sqrt[3]{x-1}-x^2-1\right)}{x^2-4x+4}\)
cao nhân nào đó giúp với , xin cảm ơn nhiều !
\(lim_{x->1}\frac{\sqrt{6-2x}-\sqrt{x^2+3}}{\left(x-1\right)^2}\)
Tính: \(\lim\limits_{x\rightarrow1}\frac{\sqrt{3x-2}+\sqrt[3]{3x+5}-\frac{7}{4}x-\frac{5}{4}}{x^2-2x+1}\)
I=\(lim_{x->1}\frac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\) Chứng minh I không tồn tại
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{6x-9}-\sqrt[3]{27x-54}}{\left(x-3\right)\left(x^2+3x-18\right)}\)
