Tìm giới hạn C = lim x → - ∞ 4 x 2 - 2 + x 3 + 1 3 x 2 + 1 - x
A. +∞
B. -∞
C. 3/2
D. 0
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}}\);
c) \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}}\).
a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} = + \infty \)
b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^+ }} \frac{{ - x}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right).\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}}\)
Ta có: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) = - \mathop {\lim }\limits_{x \to {2^ + }} x = - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} = +\infty \)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} = - \infty \)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - {1^ + }} \frac{1}{{x + 1}}\);
b) \(\mathop {\lim }\limits_{x \to - \infty } \left( {1 - {x^2}} \right)\);
c) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}\).
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: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)
\(\lim\limits_{x\rightarrow3^-}x-3=0\)
\(\lim\limits_{x\rightarrow3^-}-x=3>0\)
=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - 1} \left( {3{x^2} - x + 2} \right)\)
b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}}\)
c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}}\)
a) \(\mathop {\lim }\limits_{x \to - 1} \left( {3{x^2} - x + 2} \right) = \mathop {\lim }\limits_{x \to - 1} \left( {3{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2\)
\( = 3\mathop {\lim }\limits_{x \to - 1} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2 = 3.{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2 = 6\)
b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)
c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)
\( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)
\( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} = - \frac{1}{6}\)
a) Sử dụng phép đổi biến \(t = \frac{1}{x},\) tìm giới hạn \(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}}.\)
b) Với \(y = {\left( {1 + x} \right)^{\frac{1}{x}}},\) tính ln y và tìm giới hạn của \(\mathop {\lim }\limits_{x \to 0} \ln y.\)
c) Đặt \(t = {e^x} - 1.\) Tính x theo t và tìm giới hạn \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x}.\)
a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó
\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)
b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)
c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 7x + 4} \right)\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{{x^2} - 9}}\)
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{3 - \sqrt {x + 8} }}{{x - 1}}\)
a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)
b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)
c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)
\(=-\dfrac{1}{6}\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 3}}{{2x}}\);
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{2}{{3x + 1}}\);
c) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{{x + 1}}\).
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: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)
Tìm giới hạn hàm số Lim x->4 1-x/(x-4)^2 Lim x->3+ 2x-1/x-3 Lim x->2+ -2x+1/x+2 Lim x->1- 3x-1/x+1
1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)
2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0
3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)
\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)
4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)
Tìm giới hạn A = lim x → 1 x 3 - 3 x 2 + 2 x 2 - 4 x + 3 .
A. + ∞
B. - ∞
C. 3 2
D. 1
Tìm các giới hạn sau:
a) \(\lim \frac{{3n - 1}}{n}\)
b) \(\lim \frac{{\sqrt {{n^2} + 2} }}{n}\)
c) \(\lim \frac{2}{{3n + 1}}\)
d) \(\lim \frac{{\left( {n + 1} \right)\left( {2n + 2} \right)}}{{{n^2}}}\)
a) \(\lim \frac{{3n - 1}}{n} = \lim \frac{{n\left( {3 - \frac{1}{n}} \right)}}{n} = \lim \left( {3 - \frac{1}{n}} \right) = 3 - 0 = 3\)
b) \(\lim \frac{{\sqrt {{n^2} + 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {1 + \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {1 + \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {1 + \frac{2}{{{n^2}}}} = 1 + 0 = 1\)
c) \(\lim \frac{2}{{3n + 1}} = \lim \frac{2}{{n\left( {3 + \frac{1}{n}} \right)}} = \lim \left( {\frac{2}{n}.\frac{1}{{3 + \frac{1}{n}}}} \right) = \lim \frac{2}{n}.\lim \frac{1}{{3 + \frac{1}{n}}} = 0.\frac{1}{{3 + 0}} = 0\)
d) \(\lim \frac{{\left( {n + 1} \right)\left( {2n + 2} \right)}}{{{n^2}}} = \lim \frac{{n\left( {1 + \frac{1}{n}} \right).2n\left( {1 + \frac{1}{n}} \right)}}{{{n^2}}} = \lim \frac{{2{n^2}{{\left( {1 + \frac{1}{n}} \right)}^2}}}{{{n^2}}}\)
\( = \lim 2{\left( {1 + \frac{1}{n}} \right)^2} = 2.{\left( {1 + 0} \right)^2} = 2\)
Bài 1. Tìm các giới hạn sau:
a) \(\lim\limits\dfrac{-2n+1}{n}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)