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Question Number 194928 by cortano12 last updated on 19/Jul/23

$$\underset{x\to \mathrm{\infty }}{\lim }(\sqrt{x^2+5x+1}+\sqrt{x^2-2x+1}+\sqrt{x^2+3}+\sqrt{x^2-4x+9}-\sqrt{16x^2-8}=?$$

Answered by horsebrand11 last updated on 20/Jul/23

$$=\frac{5}{2}-\frac{2}{2}+\frac{3}{2}-\frac{4}{2}-0=\frac{2}{2}=1$$

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