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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.
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
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: \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2+5-9}{\sqrt{x^2+5}+3}\cdot\dfrac{1}{x+2}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-4}{\left(x+2\right)\left(\sqrt{x^2+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(\sqrt{x^2+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x-2}{\sqrt{x^2+5}+3}\)
\(=\dfrac{-2-2}{\sqrt{\left(-2\right)^2+5}+3}=\dfrac{-4}{3+3}=-\dfrac{4}{6}=-\dfrac{2}{3}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2+x-6}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+3x-2x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+3}{x+2}=\dfrac{2+3}{2+2}=\dfrac{5}{4}\)