\(lim_{x\rightarrow2}\frac{\left(\sqrt{x^2+6}-2x\right)\left(\sqrt{4x+1}+x\sqrt[3]{x-1}-x^2-1\right)}{x^2-4x+4}\)
cao nhân nào đó giúp với , xin cảm ơn nhiều !
\(lim_{x->1}\frac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{\left(x-1\right)^2}\)
l\(lim_{x->0}\left(1-x\right)tan\frac{\pi x}{2}\)
\(lim_{x->0}\frac{x.sin2x}{1-cos2x}\)
\(lim_{x->0}\frac{\sqrt{1-x}-1}{x}\)
\(lim_{x->0-}\frac{1}{x}\left(\frac{1}{x+1}-1\right)\)
\(lim_{x->0-}\frac{2x+\sqrt{-x}}{5x-\sqrt{-x}}\)
cho \(lim_{x->1}\dfrac{f\left(x\right)-10}{x-1}=5\) tính giới hạn \(lim_{x->1}\dfrac{f\left(x\right)-10}{\left(\sqrt{x}-1\right)\left(\sqrt[]{4f\left(x\right)+9}+3\right)}\) bằng bao nhiêu ?
\(lim_{x->\pm\infty}\sqrt{x^2-3x+4}\)
\(lim_{x->\pm\infty}x\left(\sqrt{x^2+5}+x\right)\)
\(lim_{x->2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(lim_{x\rightarrow3}\frac{\sqrt{5x+1}-2\sqrt{7x+4}+4\sqrt[3]{x+5}-x+1}{x^2-3x}\)
I=\(lim_{x->1}\frac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\) Chứng minh I không tồn tại
Các bạn tính giúp mình mấy câu này với:
1. \(\lim\limits_{x\rightarrow\left(-1\right)-}\dfrac{\sqrt{x^2-3x-4}}{1-x^2}\)
2. \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x-2}-\dfrac{x+1}{\sqrt{x+2}-2}\right)\)
3. \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x^2-5sin2x+7cos^2x}{2x^2+2}\)
4. \(\lim\limits_{x\rightarrow+\infty}\left(x.sin\left(\dfrac{1}{3x}\right)\right)\)
5. \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{2x+1}.\sqrt[3]{3x+1}.\sqrt[4]{4x+1}-1}{x}\)
6. \(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{9x+4}-\sqrt[3]{4x^{^2}+8}}{sinx}\right)\)
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-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)