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

##### 1 Answer
Aug 7, 2018

${\lim}_{x \to 1} \frac{1 - x + \ln \left(x\right)}{1 - \sqrt{2 x - {x}^{2}}} = - \infty$

#### Explanation:

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

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

So:
Using L'Hôpital's rule :

${\lim}_{x \to 1} \frac{1 - x + \ln \left(x\right)}{1 - \sqrt{2 x - {x}^{2}}} = {\lim}_{x \to 1} \frac{{\left(1 - x + \ln \left(x\right)\right)}^{'}}{{\left(1 - \sqrt{2 x - {x}^{2}}\right)}^{'}}$

$= {\lim}_{x \to 1} \frac{- 1 - \frac{1}{x}}{- \frac{1}{2} \cdot \left(2 - 2 x\right) \cdot \frac{1}{\sqrt{2 x - {x}^{2}}}}$

$= {\lim}_{x \to 1} \frac{\left(x + 1\right) \sqrt{2 x - {x}^{2}}}{x - {x}^{2}}$

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

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

${\lim}_{X \to 0} \frac{\left(X + 2\right) \sqrt{2 \left(X + 1\right) - {\left(X + 1\right)}^{2}}}{X + 1 - {\left(X + 1\right)}^{2}}$

$= {\lim}_{X \to 0} \frac{\left(X + 2\right) \sqrt{2 X + 2 - {X}^{2} - 2 X - 1}}{X + 1 - {X}^{2} - 2 X - 1}$

$= - {\lim}_{X \to 0} \frac{\left(X + 2\right) \sqrt{1 - {X}^{2}}}{{X}^{2} + X}$

$= \frac{- \left(0 + 2\right) \sqrt{1 - 0}}{0} ^ + = \frac{- 2}{0} ^ +$

$= - \infty$

\0/ Here's our answer !