a)

$\underset{x\to +\infty }{\lim }\frac{2}{x}=0$  

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b) 

$\underset{x\to -\infty }{\lim }\frac{200}{x+100}=\underset{x\to -\infty }{\lim }\frac{x\cdot \frac{200}{x}}{x\left(1+\frac{100}{x}\right)}=$

$=\underset{x\to -\infty }{\lim }\frac{\frac{200}{x}}{1+\frac{100}{x}}=\frac{0}{1}=0$   

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c)

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

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d)

$\underset{x\to +\infty }{\lim }\frac{6x^2+3}{2x^2-x}=\underset{x\to +\infty }{\lim }\frac{x^2\left(6+\frac{3}{x^2}\right)}{x^2\left(2-\frac{1}{x}\right)}=$

$=\underset{x\to +\infty }{\lim }\frac{6+\frac{3}{x^2}}{2-\frac{1}{x}}=\frac{6}{2}=3$   

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e)

$\underset{x\to +\infty }{\lim }\frac{6x+3}{2x^2-x}=\underset{x\to +\infty }{\lim }\frac{x^2\left(\frac{6}{x}+\frac{3}{x^2}\right)}{x^2\left(2-\frac{1}{x}\right)}=$

$=\underset{x\to +\infty }{\lim }\frac{\frac{6}{x}+\frac{3}{x^2}}{2-\frac{1}{x}}=\frac{0}{2}=0$   

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f)

$\underset{x\to -\infty }{\lim }\frac{6x^2+3}{2x-1}=\underset{x\to -\infty }{\lim }\frac{x^2\left(6+\frac{3}{x^2}\right)}{x\left(2-\frac{1}{x}\right)}=$

$=\underset{x\to -\infty }{\lim }\frac{x\left(6+\frac{3}{x^2}\right)}{2-\frac{1}{x}}=-\infty$   

UWAGA! Odpowiedź podana na końcu książki jest błędna.