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Question Number 141699 by cesarL last updated on 22/May/21

If lim_(x→1) =(((x)^(1/k) −1)/(x−1))=L≠0 Find lim_(x→0) (((√(x+1))−1)/( ((x+1))^(1/k) −1))

$${If}\:{lim}_{{x}\rightarrow\mathrm{1}} =\frac{\sqrt[{{k}}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}={L}\neq\mathrm{0}\:\:{Find}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$

Answered by iloveisrael last updated on 22/May/21

Given lim_(x→1) (((x)^(1/(k )) −1)/(x−1)) = L equivalent to lim_(t→1) ((t−1)/(t^k −1)) = L ⇔lim_(x→0) (((√(x+1))−1)/( ((x+1))^(1/(k )) −1)) let x+1 = t^(2k) ; x→0 ∧ t→1 lim_(t→1) ((t^k −1)/(t^2 −1)) = (1/(lim_(t→1) (((t−1)/(t^k −1))).lim_(t→1) (t+1))) = (1/(2L))

$${Given}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{{k}\:}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{L}\:{equivalent} \\ $$$${to}\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{t}−\mathrm{1}}{{t}^{{k}} −\mathrm{1}}\:=\:{L}\: \\ $$$$\Leftrightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}\:}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$$${let}\:{x}+\mathrm{1}\:=\:{t}^{\mathrm{2}{k}} \:;\:{x}\rightarrow\mathrm{0}\:\wedge\:{t}\rightarrow\mathrm{1} \\ $$$$\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{t}^{{k}} −\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{{t}−\mathrm{1}}{{t}^{{k}} −\mathrm{1}}\right).\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left({t}+\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{L}} \\ $$$$ \\ $$

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