Lời giải:
\(\lim\limits_{x\to -2}\frac{2|x-1|-5\sqrt{x^2-3}}{2x+3}=\frac{2|-2-1|-5\sqrt{(-2)^2-3}}{2.-2+3}=-1\)
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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. \(lim\frac{x+1}{\sqrt{2+x}}\left(x\rightarrow-1\right)\)
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) \(\lim\limits_{x\rightarrow-1}\dfrac{2x^3-5x-4}{\left(x+1\right)^2}\)
limlimits_{xrightarrow+infty}left(sqrt{x^2+x+1}-sqrt[3]{2x^3+x-1}right)limlimits_{xrightarrow+infty}left(sqrt{4x^2+x+1}-2xright)limlimits_{xrightarrow-infty}left(sqrt[3]{x^3+x^2+1}+sqrt{x^2+x+1}right)limlimits_{xrightarrow+infty}left(sqrt{x^2+x+1}-2sqrt{x^2-x}+xright)limlimits_{xrightarrow+infty}xleft(sqrt{x^2+2x}-2sqrt{x^2+x}+xright)
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\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-\sqrt[3]{2x^3+x-1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2+x+1}-2x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}+\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x}+x\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
limlimits_{xrightarrow-infty}left(x-sqrt{x^2+x+1}right)limlimits_{xrightarrowpminfty}left(sqrt{x^2+3x+1}-sqrt{x^2-x+1}right)limlimits_{xrightarrow+infty}left(sqrt[3]{8x^3+2x}-2xright)limlimits_{xrightarrow+infty}left(sqrt[4]{16x^4+3x+1}-sqrt{4x^2+2}right)limlimits_{xrightarrow+infty}left(sqrt{x^2+1}+sqrt{x^2-x}-2xright)
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\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow\pm\infty}\left(\sqrt{x^2+3x+1}-\sqrt{x^2-x+1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{8x^3+2x}-2x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[4]{16x^4+3x+1}-\sqrt{4x^2+2}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}+\sqrt{x^2-x}-2x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2-3x+4}-2x}{\sqrt{x^2+x+1}-x}\)
Tìm các giới hạn sau :
a) limlimits_{xrightarrow2}dfrac{x+3}{x^2+x+4}
b) limlimits_{xrightarrow-3}dfrac{x^2+5x+6}{x^2+3x}
c) limlimits_{xrightarrow4^-}dfrac{2x-5}{x-4}
d) limlimits_{xrightarrow+infty}left(-x^3+x^2-2x+1right)
e) limlimits_{xrightarrow-infty}dfrac{x+3}{3x-1}
f) limlimits_{xrightarrow-infty}dfrac{sqrt{x^2-2x+4}-x}{3x-1}
Đọc tiếp
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}\)
b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}\)
c) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}\)
d) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
e) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+3}{3x-1}\)
f) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-2x+4}-x}{3x-1}\)
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) \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
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
Tính giới hạn B = lim \(x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\) \(\left(x\rightarrow+\infty\right)\)
A. \(+\infty\)
B. \(-\infty\)
C. \(-\frac{1}{4}\)
D. 0