\(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{3x-2}}{x^4-4}=0\)
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Bài 2: Giới hạn của hàm số
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JP
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}
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\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{x-\sqrt[3]{3x+2}}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{x^2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{5+4x}-\sqrt[3]{7+6x}}{x^3+x^2-x-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+3x\right)^3-\left(1-4x\right)^4}{x}\)
\(\lim\limits_{x\rightarrow2}\dfrac{2x^2-5x+2}{x^3-3x-2}\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^4-3x+2}{x^3+2x-3}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{4-x^2}{\sqrt{2-x}}\)
Tìm các giới hạn sau:
a) \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x+1}-3}{x^2-4}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^2-4}{\sqrt{\left(x^4+1\right)\left(2-x\right)}}\)
NV
25 tháng 12 2023 lúc 21:13
Tìm \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x-4}+\sqrt{x+4}+2}{x-5}\)
Trình bài như một bài tự luận.
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow1^+}\dfrac{x^3+x+1}{x-1}\)
b) \(\lim\limits_{x\rightarrow-1^+}\dfrac{3x+2}{x+1}\)
c) \(\lim\limits_{x\rightarrow2^-}\dfrac{x-15}{x-2}\)