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Question Number 73017 by Rio Michael last updated on 05/Nov/19

$$find$$$$\underset{x\to +\mathrm{\infty }}{\lim }x\sqrt{x^2+1}$$

Commented by mathmax by abdo last updated on 05/Nov/19

$$={lim}_{x\to +\mathrm{\infty }}x\sqrt{x^2(1+\frac{1}{x^2}})={lim}_{x\to +\mathrm{\infty }}{x}^2\sqrt{1+\frac{1}{x^2}}={lim}_{x\to +\mathrm{\infty }}{x}^2=+\mathrm{\infty }$$

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