tính giới hạn
a) \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
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10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)
\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)
\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)
c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)
\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)
\(=-\sqrt{2+2}-2=-2-2=-4\)
HT
24 tháng 1 2021
a/ L'Hospital:
\(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)
b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
AH
Akai Haruma
Giáo viên
14 tháng 5 2021
Lời giải:
a)
\(\lim\limits_{x\to +\infty}\frac{\sqrt[3]{x^3+2x^2-4x+1}}{\sqrt{2x^2+x-8}}=\lim\limits_{x\to +\infty}\frac{\sqrt[3]{1+\frac{2}{x}-\frac{4}{x^2}+\frac{1}{x^3}}}{\sqrt{2+\frac{1}{x}-\frac{8}{x^2}}}\)
\(=\frac{1}{\sqrt{2}}\)
b)
\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty}\frac{\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}+1}{-3+\frac{1}{x}}=\frac{-1}{3}\)
HT
9 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}+\dfrac{3}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=\dfrac{1}{3}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{4}{x^2}}-\dfrac{x}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=-\dfrac{2}{3}\)
2
18 tháng 11 2023
`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`
AH
Akai Haruma
Giáo viên
22 tháng 2 2023
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)
\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)