ughuigyugb

a,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\lim _{x\rightarrow \ +\infty }\left(\sqrt{x^{2} -x+2} -2x\right) =\lim _{x\rightarrow \ +\infty }\left( x.\sqrt{1-\frac{1}{x} +\frac{2}{x^{2}}} -2x\right)\\
=\lim _{x\rightarrow \ +\infty }\left[ x.\left(\sqrt{1-\frac{1}{x} +\frac{2}{x^{2}}} -2\right)\right]
\end{array}$
xét $\displaystyle \lim _{x\rightarrow \ +\infty } x=+\infty $
$\displaystyle \lim _{x\rightarrow \ +\infty }\left(\sqrt{1-\frac{1}{x} +\frac{2}{x^{2}}} -2\right) =\sqrt{1-0+0} -2=-1$
vậy $\displaystyle \lim _{x\rightarrow \ +\infty }\left(\sqrt{x^{2} -x+2} -2x\right) =-\infty $

b, $\displaystyle \lim _{x\rightarrow \ 1}\frac{\sqrt{x+3} -2}{x-1} =\lim _{x\rightarrow \ 1}\frac{\left(\sqrt{x+3} -2\right) .\left(\sqrt{x+3} +2\right)}{( x-1) .\left(\sqrt{x+3} +2\right)}$
$\displaystyle \lim _{x\rightarrow 1}\frac{x+3-4}{( x-1) .\left(\sqrt{x+3} +2\right)} =\lim _{x\rightarrow \ 1}\frac{1}{\sqrt{x+3} +2} =\frac{1}{\sqrt{1+3} +2} =\frac{1}{4}$

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