Tìm các giới hạn sau: lim x → + ∞ - x 3 + x 2 - 2 x + 1
Tìm các giới hạn sau:
1/ \(\lim\limits_{x->-1}\) \(\dfrac{x^{2019}+1}{x^2+x}\)
2/ \(\lim\limits_{x->1}\) \(\dfrac{x+x^2+...+x^n-n}{x-1}\)
Lời giải:
1.
\(\lim\limits_{x\to -1}\frac{x^{2019}+1}{x^2+x}=\lim\limits_{x\to -1}\frac{(x+1)(x^{2018}-x^{2017}+x^{2016}-....-x+1)}{x(x+1)}=\lim\limits_{x\to -1}\frac{x^{2018}-x^{2017}+x^{2016}-....-x+1}{x}\)
\(=\frac{(-1)^{2018}-(-1)^{2017}+(-1)^{2016}+....-(-1)+1}{-1}\)
\(=\frac{\underbrace{1+1+....+1+1}_{2019}}{-1}=\frac{2019}{-1}=-2019\)
2.
\(\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+....+(x^n-1)}{x-1}\\ =\lim\limits_{x\to 1}\frac{(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+....+(x-1)(x^{n-1}+x^{n-2}+...+x+1)}{x-1}\)
$\lim\limits_{x\to 1}[1+(x+1)+(x^2+x+1)+....+(x^{n-1}+x^{n-2}+...+x+1)]$
$=1+2+3+....+n=n(n+1):2$
\(\)
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
Tìm giới hạn sau: \(\lim\limits_{x\rightarrow0}\dfrac{x^2-3}{x^3+x^2}\)
\(\lim\limits_{x\rightarrow0}\dfrac{x^2-3}{x^3+x^2}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow0}x^3+x^2=0^3+0^2=0\\\lim\limits_{x\rightarrow0}x^2-3=0^2-3=-3< 0\end{matrix}\right.\)
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: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) = - \mathop {\lim }\limits_{x \to {2^ + }} x = - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} = +\infty \)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} = - \infty \)
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) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ - x + 2}}{{x + 1}}\);
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{x - 2}}{{{x^2}}}\).
a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ - x + 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( { - 1 + \frac{2}{x}} \right)}}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1 + \frac{2}{x}}}{{1 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \left( { - 1} \right) + \mathop {\lim }\limits_{x \to + \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x}}} = \frac{{ - 1 + 0}}{{1 + 0}} = - 1\)
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{x - 2}}{{{x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {1 - \frac{2}{x}} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\mathop {\lim }\limits_{x \to - \infty } \left( {1 - \frac{2}{x}} \right)\)
\( = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\left( {\mathop {\lim }\limits_{x \to - \infty } 1 - \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x}} \right) = 0.\left( {1 - 0} \right) = 0\).
Tìm giới hạn sau
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+x^2}-1}{x^2}\)
\(=lim_{x->0}\left(\dfrac{1+x^2-1}{x^2\left(\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1\right)}\right)\)
\(=lim_{x->0}1=1\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{2x}}{{x - 3}}\);
b) \(\mathop {\lim }\limits_{x \to + \infty } \left( {3x - 1} \right)\).
a) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{2x}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ - }} \left( {2x} \right).\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}}\)
Ta có: \(\mathop {\lim }\limits_{x \to {3^ - }} \left( {2x} \right) = 2\mathop {\lim }\limits_{x \to {3^ - }} x = 2.3 = 6;\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}} = - \infty \)
\( \Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \frac{{2x}}{{x - 3}} = - \infty \)
b) \(\mathop {\lim }\limits_{x \to + \infty } \left( {3x - 1} \right) = \mathop {\lim }\limits_{x \to + \infty } x\left( {3 - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } x.\mathop {\lim }\limits_{x \to + \infty } \left( {3 - \frac{1}{x}} \right)\)
Ta có: \(\mathop {\lim }\limits_{x \to + \infty } x = + \infty ;\mathop {\lim }\limits_{x \to + \infty } \left( {3 - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } 3 - \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 3 - 0 = 3\)
\( \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {3x - 1} \right) = + \infty \)
Tính các giới hạn sau:
a) $\underset{x\to 3}{\mathop{\lim }}\,\left( x+2 \right);$
b) $\underset{x\to +\infty }{\mathop{\lim }}\,\left( {{x}^{2}}-x+1 \right).$