a) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\left(-6x^5+x^2\right)=\underset{x\to +\infty }{\lim }\stackrel{\stackrel{+\infty }{\uparrow }}{\overbrace{x^5}}\text{(}\stackrel{\stackrel{-6}{\uparrow }}{\overbrace{-6}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{1}{x^3}}}\text{)}=\left(+\infty \right)\cdot \left(-6\right)=-\infty$  

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b) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\left(4x^7-3x^6\right)=\underset{x\to -\infty }{\lim }\stackrel{\stackrel{-\infty }{\uparrow }}{\overbrace{x^7}}\text{(}\stackrel{\stackrel{4}{\uparrow }}{\overbrace{4}}-\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{3}{x}}}\text{)}=\left(-\infty \right)\cdot 4=-\infty$  

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c) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{3x^2-4x}{-5x^3+7}=\underset{x\to +\infty }{\lim }\frac{x^2\left(3-\frac{4}{x}\right)}{x^3\left(-5+\frac{7}{x^3}\right)}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{3}{\uparrow }}{\overbrace{3}}-\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{4}{x}}}}{\underset{\underset{+\infty }{\downarrow }}{\underbrace{x}}\text{(}\underset{\underset{-5}{\downarrow }}{\underbrace{-5}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{7}{x^3}}}\text{)}}=\frac{3}{\left(+\infty \right)\cdot \left(-5\right)}=\frac{3}{-\infty }=0$  

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d) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\frac{2x^3-9}{-4x^3+5x}=\underset{x\to -\infty }{\lim }\frac{x^3\left(2-\frac{9}{x^3}\right)}{x^3\left(-4+\frac{5}{x^2}\right)}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{2}{\uparrow }}{\overbrace{2}}-\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{9}{x^3}}}}{\underset{\underset{-4}{\downarrow }}{\underbrace{-4}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{5}{x^2}}}}=\frac{2}{-4}=-\frac{1}{2}$  

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e) Obliczamy daną granicę.

$\underset{x\to 0^{-}}{\lim }\frac{\stackrel{\stackrel{-3}{\uparrow }}{\overbrace{x^2-3}}}{\underset{\underset{0^{-}}{\downarrow }}{\underbrace{x}}}=+\infty$  

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f) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{2+{0,5}^x}{3-{\log }_{2}x}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{2}{\uparrow }}{\overbrace{2}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{{0,5}^x}}}{\underset{\underset{3}{\downarrow }}{\underbrace{3}}-\underset{\underset{+\infty }{\downarrow }}{\underbrace{{\log }_{2}x}}}=\frac{2}{3-\left(+\infty \right)}=\frac{2}{-\infty }=0$  

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g) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{3^x+5\cdot 4^x}{6^x-2\cdot 7^x}=\underset{x\to +\infty }{\lim }\frac{4^x\left({\left(\frac{3}{4}\right)}^x+5\right)}{7^x\left({\left(\frac{6}{7}\right)}^x-2\right)}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{0}{\uparrow }}{\overbrace{{\left(\frac{4}{7}\right)}^x}}\text{(}\stackrel{\stackrel{0}{\uparrow }}{\overbrace{{\left(\frac{3}{4}\right)}^x}}+\stackrel{\stackrel{5}{\uparrow }}{\overbrace{5}}\text{)}}{\underset{\underset{0}{\downarrow }}{\underbrace{{\left(\frac{6}{7}\right)}^x}}-\underset{\underset{2}{\downarrow }}{\underbrace{2}}}=\frac{0\cdot 5}{-2}=0$  

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h) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{4\cdot 8^x-2}{5\cdot 3^x+2^x}=\underset{x\to +\infty }{\lim }\frac{8^x\left(4-\frac{2}{8^x}\right)}{3^x\left(5-{\left(\frac{2}{3}\right)}^x\right)}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{+\infty }{\uparrow }}{\overbrace{{\left(\frac{8}{3}\right)}^x}}\text{(}\stackrel{\stackrel{4}{\uparrow }}{\overbrace{4}}-\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{2}{8^x}}}\text{)}}{\underset{\underset{5}{\downarrow }}{\underbrace{5}}-\underset{\underset{0}{\downarrow }}{\underbrace{{\left(\frac{2}{3}\right)}^x}}}=\frac{\left(+\infty \right)\cdot 4}{5}=\frac{+\infty }{5}=+\infty$  

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i) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{6\cdot 5^x+4\cdot 3^x}{9\cdot 4^x-2\cdot 5^x}=\underset{x\to +\infty }{\lim }\frac{5^x\left(6+4\cdot {\left(\frac{3}{5}\right)}^x\right)}{5^x\left(9\cdot {\left(\frac{4}{5}\right)}^x-2\right)}=\underset{x\to +\infty }{\lim }\frac{\stackrel{\stackrel{6}{\uparrow }}{\overbrace{6}}+4\cdot \stackrel{\stackrel{0}{\uparrow }}{\overbrace{{\left(\frac{3}{5}\right)}^x}}}{9\cdot \underset{\underset{0}{\downarrow }}{\underbrace{{\left(\frac{4}{5}\right)}^x}}-\underset{\underset{2}{\downarrow }}{\underbrace{2}}}=\frac{6}{-2}=-3$