Why is the answer -$\infty$? I'm trying to understand why the answer of this question is $-\infty$. 
The question is 
$$
\lim_{x \to 1+} \frac{x-1}{\sqrt{2x-x^2}-1}
$$
And in my last step I have $\lim_{x \to 1+} \frac{\sqrt{2x-x^2}}{1-x}$. If I plug the 1+ in the equation I get $\sqrt{2(1)-(1)^2}/(1-1)$ and so, I have $\sqrt 1/0$. Wolfram alpha says that the answer is $-\infty$. 
 A: $$\lim_{x\to 1}\frac{x-1}{\sqrt{2x-x^2}-1}=\lim_{x\to 1}\frac{x-1}{2x-x^2-1}[\sqrt{2x-x^2}+1]=-\lim_{x\to 1}\frac{x-1}{(x-1)^2}[\sqrt{2x-x^2}+1]
=-\lim_{x\to 1}\frac{\sqrt{2x-x^2}+1}{x-1}$$
Since $\lim_{x\to 1}\sqrt{2x-x^2}+1=2$ and the limit $\lim_{x\to 1}\frac{1}{x-1}$ doesn't exist, your limit doesn't exist (check this with one sided limits).
EDIT: The question was changed to calculating the limit
$$\lim_{x\to 1^+}\frac{x-1}{\sqrt{2x-x^2}-1}$$
We have 
$$\lim_{x\to 1^+}\frac{x-1}{\sqrt{2x-x^2}-1}=-\lim_{x\to 1^+}\frac{\sqrt{2x-x^2}+1}{x-1}=-(2\cdot +\infty)=-\infty$$
This is true because $x-1>0$ for $x>1$ and $\lim_{x\to 1+}x-1=0$ and thus
$$\lim_{x\to 1^+}\frac{1}{x-1}=+\infty$$
A: First you have to simplify the expression, start by multiplicating and dividing the whole expression by the conjugated of the denominator
$$ \lim_ {x \to 1^{+}} \frac{x - 1}{\sqrt{2x - x^2} - 1} = \lim_ {x \to 1^{+}} \frac{x - 1}{\sqrt{2x - x^2} - 1} \times \frac{\sqrt{2x - x^2} + 1}{\sqrt{2x - x^2} + 1} $$ $$ = \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{(\sqrt{2x - x^2})^{2} - 1^{2}} = \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{-x^2 + 2x - 1} $$
Notice that 1 is a root of the denominator, so we can factorate it using the Briot-Ruffini Method, and we get this
$$ \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{(x - 1) \times (-x + 1)} = $$ $$ \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{(-x + 1)} = $$ $$ \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{-(x -1)} = $$ $$ \lim_ {x \to 1^{+}} \frac{1}{-1} \times \frac{\sqrt{2x - x^2} + 1}{x -1} = $$ $$ -1 \times \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{x -1} $$
Now let's verify for which values of x the expression x - 1 assumes positive values
$$
x - 1 > 0 \leftrightarrow x > 1
$$
As we are aproaching to x by values greater than 1, x - 1 aproachs to 0 by positive values, so
$$
-1 \times \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{x -1} = -1 \times \frac{2}{0^{+}} = -1 \times +\infty = -\infty
$$
A: As $x \to 1^{+}$, $1-x \to 0^{-}$. Hence the limit is $1/0^{-}=-\infty$.
