Question Number 173499 by cortano1 last updated on 12/Jul/22
Commented by blackmamba last updated on 12/Jul/22
$$\:{x}=\mathrm{1}+{r}\: \\ $$$$\:{L}\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\left(\mathrm{1}+{r}\right)^{\mathrm{3}} }−\sqrt{\mathrm{3}+\left(\mathrm{1}+{r}\right)^{\mathrm{2}} }}{{r}} \\ $$$$\:\:\:\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}+\left(\mathrm{3}{r}+\mathrm{3}{r}^{\mathrm{2}} +{r}^{\mathrm{3}} \right)}−\sqrt{\mathrm{4}+\left(\mathrm{2}{r}+{r}^{\mathrm{2}} \right)}}{{r}} \\ $$$$\:\:\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\left(\frac{\mathrm{3}{r}+\mathrm{3}{r}^{\mathrm{2}} +{r}^{\mathrm{3}} }{\mathrm{8}}\right)}−\mathrm{2}\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{r}+{r}^{\mathrm{2}} }{\mathrm{4}}\right)}}{{r}} \\ $$$$\:\:\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{3}{r}+\mathrm{3}{r}^{\mathrm{2}} +{r}^{\mathrm{3}} }{\mathrm{24}}\right)−\mathrm{2}\left(\mathrm{1}+\left(\frac{\mathrm{2}{r}+{r}^{\mathrm{2}} }{\mathrm{8}}\right)\right)}{{r}} \\ $$$$\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}+\mathrm{3}{r}+{r}^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{2}+{r}}{\mathrm{4}} \\ $$$$\:\:=\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Tawa11 last updated on 13/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$