a) Obliczamy daną granicę. 

$\underset{x\to +\infty }{\lim }\left(7-\frac{2}{x}\right)=\underset{x\to +\infty }{\lim }7-\underset{x\to +\infty }{\lim }\frac{2}{x}=7-0=7$  

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b) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\left(\frac{10}{x^5}-3\right)=\underset{x\to -\infty }{\lim }\frac{10}{x^5}-\underset{x\to -\infty }{\lim }3=0-3=-3$  

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c) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\left(\frac{-3}{x^4}-\frac{7}{x^2}+\frac{8}{x}\right)=\underset{x\to +\infty }{\lim }\frac{-3}{x^4}-\underset{x\to +\infty }{\lim }\frac{7}{x^2}+\underset{x\to +\infty }{\lim }\frac{8}{x}=0-0+0=0$  

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d) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\left(\frac{5}{3x^8}+\frac{3}{4x^5}+\frac{2}{3}\right)=\underset{x\to -\infty }{\lim }\frac{5}{3x^8}-\underset{x\to -\infty }{\lim }\frac{3}{4x^5}+\underset{x\to -\infty }{\lim }\frac{2}{3}=0+0+\frac{2}{3}=\frac{2}{3}$  

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e) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\left(9-\frac{4}{x^7}+\frac{-3}{x}\right)\left(\frac{12}{x^3}-\frac{11}{x^2}-10\right)=\underset{x\to +\infty }{\lim }\left(9-\frac{4}{x^7}+\frac{-3}{x}\right)\cdot \underset{x\to +\infty }{\lim }\left(\frac{12}{x^3}-\frac{11}{x^2}-10\right)=\left(\underset{x\to +\infty }{\lim }9-\underset{x\to +\infty }{\lim }\frac{4}{x^7}+\underset{x\to +\infty }{\lim }\frac{-3}{x}\right)\cdot \left(\underset{x\to +\infty }{\lim }\frac{12}{x^3}-\underset{x\to +\infty }{\lim }\frac{11}{x^2}-\underset{x\to +\infty }{\lim }10\right)=\left(9-0+0\right)\cdot \left(0-0-10\right)=9\cdot \left(-10\right)=-90$  

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f) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\left(\frac{6}{x^7}+\frac{2}{5x^2}+1\right)\left(\frac{-3}{x^6}-\frac{9}{x^5}-\frac{1}{x}\right)=\underset{x\to -\infty }{\lim }\left(\frac{6}{x^7}+\frac{2}{5x^2}+1\right)\cdot \underset{x\to -\infty }{\lim }\left(\frac{-3}{x^6}-\frac{9}{x^5}-\frac{1}{x}\right)=\left(\underset{x\to -\infty }{\lim }\frac{6}{x^7}+\underset{x\to -\infty }{\lim }\frac{2}{5x^2}+\underset{x\to -\infty }{\lim }1\right)\cdot \left(\underset{x\to -\infty }{\lim }\frac{-3}{x^6}-\underset{x\to -\infty }{\lim }\frac{9}{x^5}-\underset{x\to -\infty }{\lim }\frac{1}{x}\right)=\left(0+0+1\right)\cdot \left(0-0-0\right)=1\cdot 0=0$  

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g) Obliczamy daną granicę.

$\underset{x\to +\infty }{\lim }\frac{7+\frac{3}{x^4}-\frac{5}{x^2}}{2-\frac{15}{x^3}-\frac{10}{3x}}=\frac{\underset{x\to +\infty }{\lim }\left(7+\frac{3}{x^4}-\frac{5}{x^2}\right)}{\underset{x\to +\infty }{\lim }\left(2-\frac{15}{x^3}-\frac{10}{3x}\right)}=\frac{\underset{x\to +\infty }{\lim }7+\underset{x\to +\infty }{\lim }\frac{3}{x^4}-\underset{x\to +\infty }{\lim }\frac{5}{x^2}}{\underset{x\to +\infty }{\lim }2-\underset{x\to +\infty }{\lim }\frac{15}{x^3}-\underset{x\to +\infty }{\lim }\frac{10}{3x}}=\frac{7+0-0}{2-0-0}=\frac{7}{2}$  

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h) Obliczamy daną granicę.

$\underset{x\to -\infty }{\lim }\frac{\frac{100}{x^{20}}-\frac{4}{3x^{10}}}{\frac{8}{5}{x}^7-\frac{1}{6}{x}^5-3}=\frac{\underset{x\to -\infty }{\lim }\left(\frac{100}{x^{20}}-\frac{4}{3x^{10}}\right)}{\underset{x\to -\infty }{\lim }\left(\frac{8}{5}{x}^7-\frac{1}{6}{x}^5-3\right)}=\frac{\underset{x\to -\infty }{\lim }\frac{100}{x^{20}}-\underset{x\to -\infty }{\lim }\frac{4}{3x^{10}}}{\underset{x\to -\infty }{\lim }\frac{8}{5}{x}^7-\underset{x\to -\infty }{\lim }\frac{1}{6}{x}^5-\underset{x\to -\infty }{\lim }3}=\frac{0-0}{0-0-3}=0$