Problems from Infinity's comparison

Development:

a) $$\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{\ln (x)\cdot (x^3-x^4)}{x-2x^5+3x^3}}=\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{-x^4\cdot \ln (x)}{-2x^5}}=\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{-\ln (x)}{-2x}}=0$$

b) $$\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{2x-x^2}{1+x}}-x]=\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{2x-x^2}{1+x}}-{\displaystyle \frac{x(1+x)}{1+x}}]=$$ $$=\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{2x-x^2-x-x^2}{1+x}}]=\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{-2x^2}{x}}=-\mathrm{\infty }$$

c) $$\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{1+x}{x}}-{\displaystyle \frac{1}{x}}{]}^{\ln (x)}=\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{1+x-1}{x}}{]}^{\ln (x)}=$$ $$=\underset{x\to +\mathrm{\infty }}{lim}{1}^{\ln (x)}=\underset{x\to +\mathrm{\infty }}{lim}1=1$$

Solution:

a) $0$

b) $-\mathrm{\infty }$

c) $1$$

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