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Question Number 91813 by jagoll last updated on 03/May/20

lim_(x→1)  ((1−x+ ln x)/(1−(√(2x−x^2 )))) ?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−{x}+\:\mathrm{ln}\:{x}}{\mathrm{1}−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}\:? \\ $$

Commented by john santu last updated on 03/May/20

L′Hopital  lim_(x→1)  ((−1+(1/x))/(−(((2−2x)/(2(√(2x−x^2 ))))))) =  lim_(x→1)  (((1−x)2(√(2x−x^2 )))/(−2x(1−x))) =   lim_(x→1)  ((−(√(2x−x^2 )))/x) = −1

$${L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\mathrm{1}+\frac{\mathrm{1}}{{x}}}{−\left(\frac{\mathrm{2}−\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−{x}\right)\mathrm{2}\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}{−\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}{{x}}\:=\:−\mathrm{1} \\ $$

Commented by john santu last updated on 03/May/20

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Commented by jagoll last updated on 03/May/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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