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a)
=
= -4.
b)
=
=
(2-x) = 4.
c)
=
=
=
=
.
d)
=
= -2.
e)
= 0 vì
(x2 + 1) =
x2( 1 +
) = +∞.
f)
=
= -∞, vì
> 0 với ∀x>0.
a) Ta có (x - 2)2 = 0 và (x - 2)2 > 0 với ∀x ≠ 2 và
(3x - 5) = 3.2 - 5 = 1 > 0.
Do đó
= +∞.
b) Ta có (x - 1) và x - 1 < 0 với ∀x < 1 và
(2x - 7) = 2.1 - 7 = -5 <0.
Do đó
= +∞.
c) Ta có (x - 1) = 0 và x - 1 > 0 với ∀x > 1 và
(2x - 7) = 2.1 - 7 = -5 < 0.
Do đó
= -∞.
a) lim= - 1/0 = - vô cùng
d) lim x(x^99-2)+1/ x(x^49-2)+1 =lim (x^99-2)/(x^49-2)=1
1/ \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}+\dfrac{x}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)
3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)
4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)
5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
\(=\dfrac{a}{n}-\dfrac{b}{m}\)
6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)
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.
\(\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+....+(x^n-1)}{x-1}\\ =\lim\limits_{x\to 1}\frac{(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+....+(x-1)(x^{n-1}+x^{n-2}+...+x+1)}{x-1}\)
$\lim\limits_{x\to 1}[1+(x+1)+(x^2+x+1)+....+(x^{n-1}+x^{n-2}+...+x+1)]$
$=1+2+3+....+n=n(n+1):2$
\(\)