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Question Number 79679 by ~blr237~ last updated on 27/Jan/20

$$Find\underset{x\to \mathrm{\infty }}{\lim }{x}^{2}ln(1+x)-x$$

Commented by mathmax by abdo last updated on 27/Jan/20

$$letf\left(x\right)=x^{2}ln(1+x)-x\Rightarrow f\left(x\right)=x^{2}ln\left(x(1+\frac{1}{x})\right)-x$$$$=x^{2}lnx+x^{2}ln(1+\frac{1}{x})-x\sim x^{2}lnx+x^2\left(\frac{1}{x}\right)-x\Rightarrow$$$$f\left(x\right)\sim x^{2}lnx\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=+\mathrm{\infty }$$$$$$

Commented by ~blr237~ last updated on 27/Jan/20

$$thankssir$$

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