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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 $-\infty$.

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    Try "plugging in" numbers such as $1.01$, $1.001$, $1.0001$, into your expression, and see what comes out.2012-12-20
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    It might be easier changing variables to $y=x-1$. Then the limit becomes $$\lim_{y\to0^+}\frac{y}{\sqrt{1-y^2}-1}$$2012-12-20
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    Thanks for the commentaries.2012-12-20
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    But in anyway may you show me the entire job?2012-12-20

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