Bài 4: Ôn tập chương Giới hạn
Các câu hỏi tương tự
Trong các giới hạn sau , giới hạn nào không tồn tại ?
A. \(lim\frac{x+1}{\sqrt{x-2}}\left(x\rightarrow1\right)\)
B. \(lim\frac{x+1}{\sqrt{-x+2}}\left(x\rightarrow-1\right)\)
C. \(lim\frac{x+1}{\sqrt{2-x}}\left(x\rightarrow1\right)\)
D. \(lim\frac{x+1}{\sqrt{2+x}}\left(x\rightarrow-1\right)\)
\(\lim\limits_{x\rightarrow0^-}\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{1}{x^3-1}-\dfrac{1}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt[3]{1-x^3}\right)\)
Mọi người giúp em với.
Tính:
a) \(\lim\limits_{x\rightarrow1}\left(\frac{x^m-1}{x^n-1}\right)\)
b) \(\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)\left(1-\sqrt[4]{x}\right)\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)^4}\)
Tìm các giới hạn sau :
a) limlimits_{xrightarrow0}dfrac{sqrt{x^2+1}-1}{4-sqrt{x^2+16}}
b) limlimits_{xrightarrow1}dfrac{x-sqrt{x}}{sqrt{x}-1}
c) limlimits_{xrightarrow+infty}dfrac{2x^4+5x-1}{1-x^2+x^4}
d) limlimits_{xrightarrow-infty}dfrac{x+sqrt{4x^2-x+1}}{1-2x}
e) limlimits_{xrightarrow+infty}xleft(sqrt{x^2+1}-xright)
f) limlimits_{xrightarrow2^+}left(dfrac{1}{x^2-4}-dfrac{1}{x-2}right)
Đọc tiếp
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2x^4+5x-1}{1-x^2+x^4}\)
d) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{4x^2-x+1}}{1-2x}\)
e) \(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
f) \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2-3x+4}-2x}{\sqrt{x^2+x+1}-x}\)
Tính :
\(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{\left(x-1\right)^2}\)
Tính giới hạn:
\(_{\lim\limits_{x\rightarrow\dfrac{\Pi}{4}}}\dfrac{\sqrt{2}Cosx-1}{\sqrt{2}Sinx-1}\)
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-2}\dfrac{x+5}{x^2+x-3}\)
b) \(\lim\limits_{x\rightarrow3^-}\sqrt{x^2+8x+3}\)
c) \(\lim\limits_{x\rightarrow+\infty}\left(x^3+2x^2\sqrt{x}-1\right)\)
d) \(\lim\limits_{x\rightarrow-1}\dfrac{2x^3-5x-4}{\left(x+1\right)^2}\)
Cho \(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-5}{x-3}=7\)
Tính \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt[3]{5f\left(x\right)-11}-4}{x^2-x-6}\)
Giúp em với ạ!!! em cảm ơn nhìu<3