Giải bài 4.55 trang 173 - SBT Đại số và Giải tích lớp 11

a)

$\begin{align} & \lim_{x\to 0}\limits\,\dfrac{\sqrt{{{x}^{2}}+1}-1}{4-\sqrt{{{x}^{2}}+16}} \\ & =\lim_{x\to 0}\limits\,\dfrac{{{x}^{2}}\left( 4+\sqrt{{{x}^{2}}+16} \right)}{\left( -{{x}^{2}} \right)\left( \sqrt{{{x}^{2}}+1}+1 \right)} \\ & =\lim_{x\to 0}\limits\,\dfrac{-\left( 4+\sqrt{{{x}^{2}}+16} \right)}{\sqrt{{{x}^{2}}+1}+1}=-4 \\ \end{align}$

b) $\lim_{x\to 1}\limits\,\dfrac{x-\sqrt{x}}{\sqrt{x}-1}=\lim_{x\to 1}\limits\,\dfrac{\sqrt{x}\left( \sqrt{x}-1 \right)}{\sqrt{x}-1}=\lim_{x\to 1}\limits\,\sqrt{x}=1$

c) $\lim_{x\to +\infty }\limits\,\dfrac{2{{x}^{4}}+5x-1}{1-{{x}^{2}}+{{x}^{4}}}=\lim_{x\to +\infty }\limits\,\dfrac{2+\dfrac{5}{{{x}^{3}}}-\dfrac{1}{{{x}^{4}}}}{\dfrac{1}{{{x}^{4}}}-\dfrac{1}{{{x}^{2}}}+1}=2$

d)
$\begin{align} & \lim_{x\to -\infty }\limits\,\dfrac{x+\sqrt{4{{x}^{2}}-x+1}}{1-2x} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{x+\left| x \right|\sqrt{4-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{x\left( \dfrac{1}{x}-2 \right)} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{1-\sqrt{4-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{\dfrac{1}{x}-2}=\dfrac{1}{2} \\ \end{align}$

e) 
$\begin{align} & \lim_{x\to +\infty }\limits\,x\left( \sqrt{{{x}^{2}}+1}-x \right) \\ & =\lim_{x\to +\infty }\limits\,\dfrac{x\left( {{x}^{2}}+1-{{x}^{2}} \right)}{\sqrt{{{x}^{2}}+1}+x} \\ & =\lim_{x\to +\infty }\limits\,\dfrac{1}{\sqrt{1+\dfrac{1}{{{x}^{2}}}}+1}=\dfrac{1}{2} \\ \end{align}$

f)
$\begin{align} & \lim_{x\to {{2}^{+}}}\limits\,\left( \dfrac{1}{{{x}^{2}}-4}-\dfrac{1}{x-2} \right) \\ & =\lim_{x\to {{2}^{+}}}\limits\,\dfrac{1-\left( x+2 \right)}{{{x}^{2}}-4} \\ & =\lim_{x\to {{2}^{+}}}\limits\,\dfrac{-x-1}{{{x}^{2}}-4}=-\infty \\ \end{align}$