Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)
\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)
\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)
\(=\frac{1^2+1+3}{1}\)
\(=5\)
\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)
\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)
\(=\frac{-1}{2+2}=\frac{-1}{4}\)
\(\lim\limits_{x\rightarrow-2}=\dfrac{x-1+\sqrt{2x^2+1}}{4-x^2}\)
\(=\lim\limits_{x\rightarrow-2}=\dfrac{\left[\left(x-1\right)+\sqrt{2x^2+1}\right]\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x-1\right)^2-\left(2x^2+1\right)}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+1-2x^2-1}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{-x^2-2x}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)
\(=\lim\limits_{x\rightarrow-2}=-\dfrac{x}{\left(2-x\right)\left(x-1-\sqrt{2x^2+1}\right)}\)
\(=-\dfrac{1}{12}\)
a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)
=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)
b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)
\(=-\infty\)
c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)
\(\lim\limits_{x\rightarrow3^-}x-3=0\)
\(\lim\limits_{x\rightarrow3^-}-x=3>0\)
=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)
a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)
b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)
c: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 = - 1\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt 1 + 1}} = \frac{1}{2}\)
a) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \frac{6}{5}\)
b) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{6 + \frac{8}{x}}}{{5 - \frac{2}{x}}} = \frac{6}{5}\).
c) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} = - \frac{3}{3} = - 1\).
d) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} = \frac{3}{3} = 1\).
e) \(\mathop {\lim }\limits_{x \to - {2^ - }} \frac{{3{x^2} + 4}}{{2x + 4}} = - \infty \)
Do \(\mathop {\lim }\limits_{x \to - {2^ - }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to - {2^ - }} \frac{1}{{2x + 4}} = - \infty \)
g) \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{{3{x^2} + 4}}{{2x + 4}} = + \infty \).
Do \(\mathop {\lim }\limits_{x \to - {2^ + }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{1}{{2x + 4}} = + \infty \)
