Dùng định nghĩa, tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow4}\dfrac{x+1}{3x-2}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2-5x^2}{x^2+3}\)
Những câu hỏi liên quan
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) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-2x+4}-x}{3x-1}\)
BÀI 3. Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^3-5x^2+1}{7x^2-x+4}\)
b) \(\lim\limits_{x\rightarrow+\infty}x\sqrt{\dfrac{x^2+2x+3}{3x^4+4x^2-5}}\)
a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)
\(=\dfrac{2x-5}{7}\)
\(=\dfrac{2}{7}x-\dfrac{5}{7}\)
\(=-\infty\)
b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
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27 tháng 1 2023 lúc 8:56
Tìm các giới hạn sau:limlimits_{xrightarrow-infty} dfrac{sqrt{x^6+2}}{3text{x}^3-1}limlimits_{xrightarrow+infty} dfrac{sqrt{x^6+2}}{3text{x}^3-1}limlimits_{xrightarrow-infty} left(sqrt{2text{x}^2+1}+xright)limlimits_{xrightarrow1} dfrac{2text{x}^3-5text{x}-4}{left(x+1right)^2}
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Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x+1}{x^2+x+1}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x+1}{3x^2-x+5}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x+5}{\sqrt{x^2+x}}\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{-5x+1}{\sqrt{3x^2+1}}\)
`a)lim_{x->+oo}[x+1]/[x^2+x+1]`
`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`
`=0`
`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`
`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`
`=0`
`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`
`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`
`=-3`
`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`
`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`
`=-5/3`
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Tính các giới hạn :
a) limlimits_{xrightarrow+infty}left(dfrac{x^3}{3x^2-4}-dfrac{x^2}{3x+2}right)
b) limlimits_{xrightarrow+infty}left(sqrt{9x^2+1}-3xright)
c) limlimits_{xrightarrow-infty}left(sqrt{2x^2-3}-5xright)
d) limlimits_{xrightarrow+infty}dfrac{sqrt{2x^2+3}}{4x+2}
e) limlimits_{xrightarrow-infty}dfrac{sqrt{2x^2+3}}{4x+2}
Đọc tiếp
Tính các giới hạn :
a) \(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x^3}{3x^2-4}-\dfrac{x^2}{3x+2}\right)\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{9x^2+1}-3x\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{2x^2-3}-5x\right)\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)e) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
a/ \(=\lim\limits_{x\rightarrow4^-}\dfrac{5-2x}{4-x}=\dfrac{-3}{0}=-\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{2}{x^2}+\dfrac{1}{x^3}\right)=-\infty\)
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Tính giới hạn
a) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}=-\infty\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
a/ \(=\lim\limits_{x\rightarrow4^-}\dfrac{5-2x}{4-x}=-\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}x^3\left(-1\right)=-\infty\)
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Dùng định nghĩa tìm các giới hạn :
a) \(\lim\limits_{x\rightarrow5}\dfrac{x+3}{3-x}\)\(\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x^3+1}{x^2+1}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]` `ĐK: 4x^3+1 ne 0`
`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`
`=1/4`
`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]` `ĐK: x ne 0;x ne 1`
`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`
`=0`
Câu `c` giống `b`.
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