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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.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) 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: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)
\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)
\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)
c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)
\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)
\(=-\sqrt{2+2}-2=-2-2=-4\)
\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt[3]{2x+12}+x}{x^2+2x}\)
\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{2x+12+x^3}{\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2}\cdot\dfrac{1}{x^2+2x}\right)\)
\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+2x^2-2x^2-4x+6x+12}{\left(\sqrt[3]{\left(2x+12\right)^2}+x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\right)\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+6\right)}{\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+6}{x\cdot\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)}\)
\(=\dfrac{\left(-2\right)^2-2\cdot\left(-2\right)+6}{\sqrt[3]{\left(-2\cdot2+12\right)^2}-\left(-2\right)\cdot\sqrt[3]{2\cdot\left(-2\right)+12}+\left(-2\right)^2}\)
\(=\dfrac{4+4+6}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}\)
\(=\dfrac{14}{8+2\cdot2+4}=\dfrac{14}{12+4}=\dfrac{14}{16}=\dfrac{7}{8}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{x^3-8}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x^2-x-2}{x+\sqrt{x+2}}\cdot\dfrac{1}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+\sqrt{x+2}\right)\cdot\left(x-2\right)\left(x^2+2x+4\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+1}{\left(x+\sqrt{x+2}\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{2+1}{\left(2+\sqrt{2+2}\right)\left(2^2+2\cdot2+4\right)}\)
\(=\dfrac{3}{\left(2+2\right)\left(4+4+4\right)}=\dfrac{3}{12\cdot4}=\dfrac{1}{4\cdot4}=\dfrac{1}{16}\)