Đáp án A, khi \(x\rightarrow1\) thì \(x-2< 0\) nên biểu thức không xác định
\(\Rightarrow\) Giới hạn đã cho ko tồn tại
Đáp án A, khi \(x\rightarrow1\) thì \(x-2< 0\) nên biểu thức không xác định
\(\Rightarrow\) Giới hạn đã cho ko tồn tại
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) \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
\(\lim\limits_{x\rightarrow0^-}\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{1}{x^3-1}-\dfrac{1}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt[3]{1-x^3}\right)\)
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) \(\lim\limits_{x\rightarrow-2}\dfrac{x+5}{x^2+x-3}\)
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) \(\lim\limits_{x\rightarrow-1}\dfrac{2x^3-5x-4}{\left(x+1\right)^2}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow\pm\infty}\left(\sqrt{x^2+3x+1}-\sqrt{x^2-x+1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{8x^3+2x}-2x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[4]{16x^4+3x+1}-\sqrt{4x^2+2}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}+\sqrt{x^2-x}-2x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-\sqrt[3]{2x^3+x-1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2+x+1}-2x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}+\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x}+x\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-x+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}x\left(\sqrt{4x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(4x^5-3x^3+x+1\right)\)
\(\lim\limits_{x\rightarrow+\infty}\sqrt{x^4-x^3+x^2-x}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2-3x+4}-2x}{\sqrt{x^2+x+1}-x}\)
Tìm giới hạn D = lim \(\left(\sqrt[3]{x^3+x^2+1}+\sqrt{x^2+x+1}\right)\) \(\left(x\rightarrow-\infty\right)\)
A. \(+\infty\)
B. \(-\infty\)
C. \(-\frac{1}{6}\)
D. 0