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Question-169803




Question Number 169803 by cortano1 last updated on 09/May/22
Answered by greougoury555 last updated on 09/May/22
 x−1= y    lim_(y→0)  (((y+1)−((((n+1)(y+1)^n −1)/n))^(1/(n+1)) )/y^(2 ) )   = lim_(y→0)  ((1−(((((n(n+1))/n)(y+1)^(n−1) )((((n+1)(y+1)^n −1)/n))^((−n)/(n+1)) )/(n+1)))/(2y))
$$\:{x}−\mathrm{1}=\:{y}\: \\ $$$$\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({y}+\mathrm{1}\right)−\left(\frac{\left({n}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }{{y}^{\mathrm{2}\:} }\: \\ $$$$=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{\left(\frac{{n}\left({n}+\mathrm{1}\right)}{{n}}\left({y}+\mathrm{1}\right)^{{n}−\mathrm{1}} \right)\left(\frac{\left({n}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}}\right)^{\frac{−{n}}{{n}+\mathrm{1}}} }{{n}+\mathrm{1}}}{\mathrm{2}{y}} \\ $$$$ \\ $$

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