$\text{a)}$  

$\underset{x\to +\infty }{\lim }\sqrt{4x^2+x-1}=\underset{x\to +\infty }{\lim }\sqrt{x^2\left(4+\frac{1}{x}-\frac{1}{x^2}\right)}=\underset{x\to +\infty }{\lim }\left(\sqrt{x^2}\cdot \sqrt{4+\frac{1}{x}-\frac{1}{x^2}}\right)=$  

$=\underset{x\to +\infty }{\lim }\left(\left|x\right|\cdot \sqrt{4+\frac{1}{x}-\frac{1}{x^2}}\right)\stackrel{\left(\ast \right)}{=}\underset{x\to +\infty }{\lim }\left(x\cdot \sqrt{4+\frac{1}{x}-\frac{1}{x^2}}\right)=$  

$=\left[+\infty \cdot \sqrt{4}\right]=+\infty$  

$\left(\ast \right)$  Ponieważ  $x\to +\infty ,$  to  $\left|x\right|=x.$  

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$\text{b)}$  

$\underset{x\to -\infty }{\lim }\left(x\sqrt{3x^2-2x+4}\right)=\underset{x\to -\infty }{\lim }(x\sqrt{x^2\left(3-\frac{2}{x}+\frac{4}{x^2}\right)}=$  

$=\underset{x\to -\infty }{\lim }\left(x\cdot \sqrt{x^2}\cdot \sqrt{3-\frac{2}{x}+\frac{4}{x^2}}\right)=\underset{x\to -\infty }{\lim }\left(x\cdot \left|x\right|\cdot \sqrt{3-\frac{2}{x}+\frac{4}{x^2}}\right)\stackrel{\left(\ast \right)}{=}$  

$=\underset{x\to -\infty }{\lim }\left(x\cdot \left(-x\right)\cdot \sqrt{3-\frac{2}{x}+\frac{4}{x^2}}\right)=\underset{x\to -\infty }{\lim }\left(-x^2\cdot \sqrt{3-\frac{2}{x}+\frac{4}{x^2}}\right)=$  

$=\left[-\infty \cdot \sqrt{3}\right]=-\infty$  

$\left(\ast \right)$  Ponieważ  $x\to -\infty ,$  to  $\left|x\right|=-x.$  

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$\text{c)}$  

$\underset{x\to -\infty }{\lim }\frac{2x\sqrt{x^2+5}}{7x^2+2}=\underset{x\to -\infty }{\lim }\frac{2x\sqrt{x^2\left(1+\frac{5}{x^2}\right)}}{x^2\left(7+\frac{2}{x^2}\right)}=\underset{x\to -\infty }{\lim }\frac{2x\cdot \sqrt{x^2}\cdot \sqrt{1+\frac{5}{x^2}}}{x^2\left(7+\frac{2}{x^2}\right)}=$  

$=\underset{x\to -\infty }{\lim }\frac{2x\cdot \left|x\right|\cdot \sqrt{1+\frac{5}{x^2}}}{x^2\left(7+\frac{2}{x^2}\right)}\stackrel{\left(\ast \right)}{=}\underset{x\to -\infty }{\lim }\frac{2x\cdot \left(-x\right)\cdot \sqrt{1+\frac{5}{x^2}}}{x^2\left(7+\frac{2}{x^2}\right)}=$  

$=\underset{x\to -\infty }{\lim }\frac{-2x^2\cdot \sqrt{1+\frac{5}{x^2}}}{x^2\left(7+\frac{2}{x^2}\right)}=\underset{x\to -\infty }{\lim }\frac{-2\sqrt{1+\frac{5}{x^2}}}{7+\frac{2}{x^2}}=-\frac{2}{7}$  

$\left(\ast \right)$  Ponieważ  $x\to -\infty ,$  to  $\left|x\right|=-x.$  

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$\text{d)}$  

$\underset{x\to +\infty }{\lim }\frac{8x\sqrt{2x^2-3}}{4x^2+2x-1}=\underset{x\to +\infty }{\lim }\frac{8x\sqrt{x^2\left(2-\frac{3}{x^2}\right)}}{x^2\left(4+\frac{2}{x}-\frac{1}{x^2}\right)}=\underset{x\to +\infty }{\lim }\frac{8x\cdot \sqrt{x^2}\cdot \sqrt{2-\frac{3}{x^2}}}{x^2\left(4+\frac{2}{x}-\frac{1}{x^2}\right)}=$  

$=\underset{x\to +\infty }{\lim }\frac{8x\cdot \left|x\right|\cdot \sqrt{2-\frac{3}{x^2}}}{x^2\left(4+\frac{2}{x}-\frac{1}{x^2}\right)}\stackrel{\left(\ast \right)}{=}\underset{x\to +\infty }{\lim }\frac{8x\cdot x\cdot \sqrt{2-\frac{3}{x^2}}}{x^2\left(4+\frac{2}{x}-\frac{1}{x^2}\right)}=\underset{x\to +\infty }{\lim }\frac{8x^2\cdot \sqrt{2-\frac{3}{x^2}}}{x^2\left(4+\frac{2}{x}-\frac{1}{x^2}\right)}=$  

$=\underset{x\to +\infty }{\lim }\frac{8\sqrt{2-\frac{3}{x^2}}}{4+\frac{2}{x}-\frac{1}{x^2}}=\frac{8\sqrt{2}}{4}=2\sqrt{2}$  

$\left(\ast \right)$  Ponieważ  $x\to +\infty ,$  to  $\left|x\right|=x.$