Question Number 193976 by Risandu last updated on 25/Jun/23
Answered by Subhi last updated on 25/Jun/23
$$ \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \frac{\left(^{\mathrm{3}} \sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \frac{\left(^{\mathrm{3}} \sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }{\left(^{\mathrm{3}} \sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} \left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+^{\mathrm{3}} \sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rrightarrow\:{lim}_{{x}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+^{\mathrm{3}} \sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${or} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \frac{\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{−\mathrm{1}}{\mathrm{3}}} −\frac{\mathrm{2}}{\mathrm{3}}{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{2}\left({x}−\mathrm{1}\right)}\:\:\left\{{L}\:{Hopital}'{s}\:{law}\right\} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \frac{\frac{−\mathrm{2}}{\mathrm{9}}{x}^{\frac{−\mathrm{4}}{\mathrm{3}}} +\frac{\mathrm{4}}{\mathrm{9}}{x}^{\frac{−\mathrm{5}}{\mathrm{3}}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$
Answered by cortano12 last updated on 25/Jun/23
![L = [ lim_(x→1) (((x)^(1/3) −1)/(x−1)) ]^2 L = [ lim_(x→1) (1/(((x)^(1/3) )^2 +(x)^(1/3) +1)) ]^2 L = determinant (((1/9)))](https://www.tinkutara.com/question/Q193979.png)
$$\:\:\:\mathrm{L}\:=\:\left[\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\right]^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{L}\:=\:\left[\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{x}}\:\right)^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\mathrm{x}}\:+\mathrm{1}}\:\right]^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{L}\:=\:\begin{array}{|c|}{\frac{\mathrm{1}}{\mathrm{9}}}\\\hline\end{array}\: \\ $$$$ \\ $$