a) 

$\underset{x\to \infty }{\lim }\frac{6x^2+3x-4}{-2x^2+2x}=\underset{x\to \infty }{\lim }\frac{x^2\left(6+\frac{3}{x}-\frac{4}{x^2}\right)}{x^2\left(-2+\frac{2}{x}\right)}=$

$=\underset{x\to \infty }{\lim }\frac{6+\frac{3}{x}-\frac{4}{x^2}}{-2+\frac{2}{x}}=\frac{6+0-0}{-2+0}=\frac{6}{-2}=-3$   

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b)

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

$=\underset{x\to \infty }{\lim }\frac{2-\frac{3}{x^3}}{4+\frac{1}{x}-\frac{1}{x^4}}=\frac{2-0}{4+0-0}=\frac{2}{4}=\frac{1}{2}$   

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c)

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

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

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d)

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

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

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e)

$\underset{x\to \infty }{\lim }\left(x^3-8x^2-9x\right)=\underset{x\to \infty }{\lim }{x}^3\left(1-\frac{8}{x}-\frac{9}{x^2}\right)=$  

$=\infty \cdot \left(1-0-0\right)=\infty$  

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f)

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

$=-\infty \cdot \left(2+0+0\right)=-\infty$