Giá trị lim x → - 1 x 2 - 1 x + 1 bằng
A. 2
B. 1
C. 0
D. -2
Giá trị của các giới hạn :
a, lim\(\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+1}\right)\) khi x→\(-\infty\)
b, lim\(\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\) khi x→\(+\infty\)
c, lim\(\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\) khi x→\(+\infty\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)
Check lai ho minh nhe :v
giá trị của \(\lim\limits_{x\to -∞} f(x)=\dfrac{2x-1}{\sqrt{x^2+1}-1}\)
\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)\)
=\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x-1}{\sqrt{x^2+1}-1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(2-\dfrac{1}{x}\right)}{-x\cdot\sqrt{1+\dfrac{1}{x^2}}-1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2-\dfrac{1}{x}}{-\sqrt{1+\dfrac{1}{x^2}}-\dfrac{1}{x}}=\dfrac{2-0}{-\sqrt{1+0}-0}=\dfrac{2}{-1}=-2\)
cho hàm số f(x) thỏa mãn: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=2\) và \(\lim\limits_{x\rightarrow1^-}f\left(x\right)=2\). tính giá trị \(\lim\limits_{x\rightarrow1}f\left(x\right)=?\)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=2\)
cho biết \(\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{4x^2-x+5}}{a\left|x\right|+2}=\dfrac{2}{3}\). tính giá trị a?
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{4x^2-x+5}}{-ax+2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}+\sqrt{4-\dfrac{1}{x}+\dfrac{5}{x^2}}}{-a+\dfrac{2}{x}}=\dfrac{2}{-a}=\dfrac{2}{3}\)
\(\Rightarrow a=-3\)
Tính giá trị giới hạn lim (x → 0) \(\dfrac{\left(x^2+\Pi^{21}\right)\sqrt[7]{1-2x}-\Pi^{21}}{x}\) là:
Tính giá trị giới hạn \(lim\left(x\rightarrow0\right)\dfrac{\left(x^2+\pi^{21}\right)\sqrt[7]{1-2x}-\pi^{21}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+\pi^{21}\right)\left(1-2x\right)^{\dfrac{1}{7}}-\pi^{21}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{7}\left(1-2x\right)^{-\dfrac{6}{7}}.\left(-2\right)\left(x^2+\pi^{21}\right)+2x\left(1-2x\right)^{\dfrac{1}{7}}}{1}\)
\(=\dfrac{1}{7}.\left(-2\right).\pi^{21}=...\)
Cho f ( x ) = x 2018 = 1009 x 2 + 2019 x Giá trị của lim ∆ x → 0 f ( ∆ x + 1 ) - f ( 1 ) ∆ x bằng
A. 1009
B. 1008
C. 2018
D. 2019
giá trị của \(\lim\limits_{x\to -∞} f(x)=\dfrac{\sqrt{x^2-3}}{x+3}\)
\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-3}}{x+3}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2\left(1-\dfrac{3}{x^2}\right)}}{x+3}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\cdot\sqrt{1-\dfrac{3}{x^2}}}{x\left(1+\dfrac{3}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{1-\dfrac{3}{x^2}}}{1+\dfrac{3}{x}}\)
\(=\dfrac{-\sqrt{1-0}}{1+0}=-\dfrac{1}{1}=-1\)
Cho f x = x 2018 = 1009 x 2 + 2019 x . Giá trị của lim △ x → 0 f △ x + 1 - f 1 △ x bằng:
A. 1009
B. 1008
C. 2018
D. 2019
Cho lim ( \(\sqrt{x^2+ax+5}+x\)) =5 Giá trị của a bằng bao nhiêu ?
x-> -∞
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+ax+5-x^2}{\sqrt{x^2+ax+5}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{ax}{x}+\dfrac{5}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{ax}{x^2}+\dfrac{5}{x^2}}-\dfrac{x}{x}}=\dfrac{-a}{2}\)
\(-\dfrac{a}{2}=5\Rightarrow a=-10\)