\(=\lim\limits_{x\rightarrow0}\dfrac{x^2\left(x^6-3x^4+5x^2+1\right)}{x^2\left(3x^8-7x^3+5\right)}=\lim\limits_{x\rightarrow0}\dfrac{x^6-3x^4+5x^2+1}{3x^8-7x^3+5}=\dfrac{0-0+0+1}{0-0+5}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x^2\left(x^6-3x^4+5x^2+1\right)}{x^2\left(3x^8-7x^3+5\right)}=\lim\limits_{x\rightarrow0}\dfrac{x^6-3x^4+5x^2+1}{3x^8-7x^3+5}=\dfrac{0-0+0+1}{0-0+5}\)
Tính giới hạn
a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}=\dfrac{1}{2}\)
b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}=\dfrac{1}{3}\)
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. \(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{9x+4}-\sqrt[3]{4x^{^2}+8}}{sinx}\right)\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}=\dfrac{1}{2}\)
b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}=\dfrac{1}{3}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+5x}+\sqrt{4x^2-x}+3x}{\sqrt{4x^2-7x}+2x}\)
Tính giới hạn: \(A=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\sqrt[5]{1-5x}-2017}{x}\)
tính \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{x^2}\)
\\(\\lim\\limits_{x\\rightarrow-\\infty}\\left(2x^3-x^2+3x-5\\right)\\)
\n\n\\(\\lim\\limits_{x\\rightarrow2}\\frac{3}{\\left(x-2\\right)\\left(x^2-3x+2\\right)}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow0}\\frac{x^2-5}{x^5+x^4}\\)
\nTìm giới hạn: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[4]{1+4x}-1}{x}\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow0^-}\dfrac{2\left|x\right|+x}{x^2-x}\)
b) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)