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Question Number 215831 by golsendro last updated on 19/Jan/25

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{{(\mathrm{x}+1)}^3}-\sqrt{{(\mathrm{x}-1)}^3}}{\sqrt{\mathrm{x}}}=?$$

Answered by efronzo1 last updated on 19/Jan/25

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{{6\mathrm{x}}^2+2}{\sqrt{\mathrm{x}{(\mathrm{x}+1)}^3}+\sqrt{\mathrm{x}{(\mathrm{x}-1)}^3}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^2(6+\frac{2}{{\mathrm{x}}^2})}{{\mathrm{x}}^2(\sqrt{1+\frac{3}{\mathrm{x}}+..}+\sqrt{1-\frac{3}{\mathrm{x}}+...})}$$$$=\frac{6}{1+1}=3$$

Answered by MrGaster last updated on 19/Jan/25

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{{(x+1)}^{\frac{3}{2}}-{(x-1)}^{\frac{3}{2}}}{\sqrt{\pi }}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{(x^{\frac{3}{2}}+\frac{3}{2}{x}^{\frac{1}{2}})-(x^{\frac{3}{2}}-\frac{3}{2}{x}^{\frac{1}{2}})}{\sqrt{\pi }}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{3\sqrt{x}}{\sqrt{x}}$$$$=\underset{x\to \mathrm{\infty }}{\lim 3}$$$$=\overline{)\begin{array}{c}3\end{array}}$$

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