Question Number 175137 by cortano1 last updated on 20/Aug/22
$$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\:\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\:\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\:\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}}=? \\ $$
Answered by Ar Brandon last updated on 20/Aug/22
![L=lim_(x→2) ((((5x−9))^(1/5) ((4x−7))^(1/4) ((3x−5))^(1/3) (√(2x−3))−1)/(x−2)) =lim_(t→0) (((((5t+1))^(1/5) )(((4t+1))^(1/4) )(((3t+1))^(1/3) )((√(2t+1)))−1)/t) =lim_(t→0) (1/t)[(1+t)(1+t)(1+t)(1+t)−1] =lim_(t→0) (1/t)[(t^2 +2t+1)^2 −1] =lim_(t→0) (1/t)(1+2(t^2 +2t)−1) =lim_(t→0) (1/t)(2t^2 +4t)=4](Q175138.png)
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\sqrt[{\mathrm{5}}]{\mathrm{5}{t}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{4}{t}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{3}{t}+\mathrm{1}}\right)\left(\sqrt{\mathrm{2}{t}+\mathrm{1}}\right)−\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{t}}\left[\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}\right)−\mathrm{1}\right] \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{t}}\left[\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{t}}\left(\mathrm{1}+\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{2}{t}\right)−\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{t}}\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}\right)=\mathrm{4} \\ $$