Đây ko phải dạng vô định nên bạn cứ thay số là được
\(=\frac{-3}{0}=-\infty\)
Chương 4: GIỚI HẠN
Các câu hỏi tương tự
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-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)
\\(\\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\\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}\\)
\n
limu_nleft(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)...left(1-frac{1}{n^2}right)
nrightarrowinftylimu_nnleft(sqrt[3]{n^{alpha}+1}-nright) nrightarrowinftylimfrac{sqrt[n]{x}-1}{sqrt[m]{x}-1} xrightarrow1limfrac{sqrt{2}-2cos x}{4x-pi} xrightarrowfrac{pi}{4}limleft(sinfrac{1}{x}+cosfrac{1}{x}right)^x
xrightarrowinftylimfrac{x^{n+1}-left(n+1right)x+n}{left(x-1right)^2} xrightarrow1limleft(1-xright)tanfrac{pi x}{2} xrightarrow1limfrac{sqrt{cos x}-sqrt[3]{cos x}}{sin^2x} xrightarrow0limsqrt[x]{1-2...
Đọc tiếp
\(limu_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{n^2}\right)\\
n\rightarrow\infty\)\(limu_n=n\left(\sqrt[3]{n^{\alpha}+1}-n\right)\\ n\rightarrow\infty\)\(lim\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}\\ x\rightarrow1\)\(lim\frac{\sqrt{2}-2\cos x}{4x-\pi}\\ x\rightarrow\frac{\pi}{4}\)\(lim\left(\sin\frac{1}{x}+\cos\frac{1}{x}\right)^x\\
x\rightarrow\infty\)\(lim\frac{x^{n+1}-\left(n+1\right)x+n}{\left(x-1\right)^2}\\ x\rightarrow1\)\(lim\left(1-x\right)\tan\frac{\pi x}{2}\\ x\rightarrow1\)\(lim\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}\\ x\rightarrow0\)\(lim\sqrt[x]{1-2x}\\ x\rightarrow0\)\(lim\left(\sin x\right)^{\tan x}\\ x\rightarrow\frac{\pi}{2}\)
\(\lim \frac{\left( x^{5}-5x^{3}+2x^{2}+6x-4\right) }{\left( x^{3}-x^{2}-x+1\right) }\)
\(\lim\limits_{x\rightarrow4}\frac{2x-\sqrt{3x+1}}{x^2-1}\)
\(\lim\limits_{x\rightarrow8}\frac{\sqrt[3]{x}-\sqrt{x-4}}{x-8}\)
\(\lim\limits_{x\rightarrow1}\frac{x-x^2}{\left(2x-1\right)\left(x^5-3\right)}\)
\(\lim\limits_{x\rightarrow0}x\left(1-\frac{1}{x}\right)\)
Tính giới hạn
a) \(\lim\limits_{x->0}\dfrac{\sqrt[m]{2x+1}-1}{\sqrt[n]{x+1}-1}\)
b) \(\lim\limits_{x->3}\dfrac{\sqrt[4]{5x+1}-2}{x-3}\)
Tìm giới hạn : \(\lim\limits_{x\rightarrow2}\frac{x^2-5x+6}{x^3-x^2-x-2}\)