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Question Number 173499 by cortano1 last updated on 12/Jul/22

Commented by blackmamba last updated on 12/Jul/22

$$x=1+r$$$$L=\underset{r\to 0}{\lim }\frac{\sqrt[3]{7+{(1+r)}^3}-\sqrt{3+{(1+r)}^2}}{r}$$$$=\underset{r\to 0}{\lim }\frac{\sqrt[3]{8+(3r+3r^2+r^3)}-\sqrt{4+(2r+r^2)}}{r}$$$$=\underset{r\to 0}{\lim }\frac{2\sqrt[3]{1+\left(\frac{3r+3r^2+r^3}{8}\right)}-2\sqrt{1+\left(\frac{2r+r^2}{4}\right)}}{r}$$$$=\underset{r\to 0}{\lim }\frac{2(1+\frac{3r+3r^2+r^3}{24})-2(1+\left(\frac{2r+r^2}{8}\right))}{r}$$$$=\underset{r\to 0}{\lim }\frac{3+3r+r^2}{12}-\frac{2+r}{4}$$$$=-\frac{1}{4}$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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