How do you find the limit of $(1 - x + log_ex) / (1 - sqrt(2x - x^2))$ as x approaches 1?

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Answer 1 · 2018-08-07T01:28:29

$lim_(x to 1)(1-x+ln(x))/(1-sqrt(2x-x^2))=-oo$

$lim_(x to 1)(1-x+ln(x))/(1-sqrt(2x-x^2)$

Both functions $f(x)=1-x+ln(x)$ and $g(x)=1-sqrt(2x-x^2)$ are defined and continuous on 1.

So:
Using L'Hôpital's rule :

$lim_(x to 1)(1-x+ln(x))/(1-sqrt(2x-x^2))=lim_(x to 1)((1-x+ln(x))^')/((1-sqrt(2x-x^2))^')$

$=lim_(x to 1)(-1-1/x)/(-1/2*(2-2x)*1/sqrt(2x-x^2))$

$=lim_(x to 1)((x+1)sqrt(2x-x^2))/(x-x^2)$

Note, here you can already see that the $f to oo$ , but you can't know the sign. So, we need the last step to know it.

Let $X=x-1$ , we have :

$lim_(X to 0) ((X+2)sqrt(2(X+1)-(X+1)^2))/(X+1-(X+1)^2)$

$=lim_(X to 0)((X+2)sqrt(2X+2-X^2-2X-1))/(X+1-X^2-2X-1)$

$=-lim_(X to 0)((X+2)sqrt(1-X^2))/(X^2+X)$

$=(-(0+2)sqrt(1-0))/0^+=(-2)/0^+$

$=-oo$

\0/ Here's our answer !

How do you find the limit of (1 - x + log_ex) / (1 - sqrt(2x - x^2)) (1 - x + log_ex) / (1 - sqrt(2x - x^2)) as x approaches 1?