\(=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{\left(2-x\right)^2}}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{2-x}}{\sqrt{x^2+1}}=\frac{0}{\sqrt{5}}=0\)
Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
\(lim_{x\rightarrow2^-}\left(\dfrac{1}{x-2}-\dfrac{1}{x^2-4}\right)\)
\(lim_{x\rightarrow2^-}\frac{\left|x-2\right|}{x-2}\)
\(lim_{x\rightarrow2}\dfrac{3x-5}{\left(x-2\right)^2}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^2-4}{\sqrt{\left(x^4+1\right)\left(2-x\right)}}\)
\(lim_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\left(\sqrt{\dfrac{3x}{x^2-1}}\right)\)
7 tháng 11 2023 lúc 19:44
limlimits_{xrightarrow2}left(dfrac{1}{x-2}-dfrac{12}{x^3-8}right)limlimits_{xrightarrow2}left(dfrac{1}{x^2-3x+2}+dfrac{1}{x^2-5x+6}right)limlimits_{xrightarrow+infty}xleft(sqrt{x^2+1}-xright)limlimits_{xrightarrow+infty}left(sqrt{x^2+1}-sqrt[3]{x^3-1}right)limlimits_{xrightarrow0}dfrac{sqrt{x^2+1}-1}{sqrt{x^2+16}-4}
Đọc tiếp
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
Tính các giới hạn sau đây :
L_1limfrac{x^3+3x^2-2x}{x^5+4x}left(xrightarrow0right)
L_2limfrac{x^3-3x+2}{left(4-2xright)^3}left(xrightarrow+inftyright)
L_3limfrac{2x^2+3x+1}{x^2+x}left(xrightarrow-1right)
L_4limfrac{x^2-4x+1}{4-x^2}left(xrightarrow2right)
L_5limfrac{sqrt{x+1}-2}{x-2}left(xrightarrow3right)
L_6limfrac{sqrt{x+3}-x-1}{x^2-1}left(xrightarrow1right)
L_7limleft(sqrt{x^2+x+1}-x+1right)left(xrightarrow+inftyright)
L_8limleft(sqrt{x^2+x+1}-3x+2right)left(xrightarrow-...
Đọc tiếp
Tính các giới hạn sau đây :
\(L_1=lim\frac{x^3+3x^2-2x}{x^5+4x}\left(x\rightarrow0\right)\)
\(L_2=lim\frac{x^3-3x+2}{\left(4-2x\right)^3}\left(x\rightarrow+\infty\right)\)
\(L_3=lim\frac{2x^2+3x+1}{x^2+x}\left(x\rightarrow-1\right)\)
\(L_4=lim\frac{x^2-4x+1}{4-x^2}\left(x\rightarrow2\right)\)
\(L_5=lim\frac{\sqrt{x+1}-2}{x-2}\left(x\rightarrow3\right)\)
\(L_6=lim\frac{\sqrt{x+3}-x-1}{x^2-1}\left(x\rightarrow1\right)\)
\(L_7=lim\left(\sqrt{x^2+x+1}-x+1\right)\left(x\rightarrow+\infty\right)\)
\(L_8=lim\left(\sqrt{x^2+x+1}-3x+2\right)\left(x\rightarrow-\infty\right)\)