Why is the answer -$\infty$?

Question

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 $-\infty$.

Try "plugging in" numbers such as $1.01$, $1.001$, $1.0001$, into your expression, and see what comes out. — David Mitra, Dec 20 '12 at 14:06
It might be easier changing variables to $y=x-1$. Then the limit becomes $$\lim_{y\to0^+}\frac{y}{\sqrt{1-y^2}-1}$$ — yohBS, Dec 20 '12 at 15:26

Nameless · Answer 1 · 2012-12-20T14:17:59.743

5

$$\lim_{x\to 1}\frac{x-1}{\sqrt{2x-x^2}-1}=\lim_{x\to 1}\frac{x-1}{2x-x^2-1}[\sqrt{2x-x^2}+1]=-\lim_{x\to 1}\frac{x-1}{(x-1)^2}[\sqrt{2x-x^2}+1] =-\lim_{x\to 1}\frac{\sqrt{2x-x^2}+1}{x-1}$$ 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 $$\lim_{x\to 1^+}\frac{1}{x-1}=+\infty$$

edited Dec 20 '12 at 14:17

answered Dec 20 '12 at 14:03

Nameless

13,456

1

Thanks a lot, in the end a I realize what i have missed. – Vinicius L. Beserra Dec 20 '12 at 18:25
@ViniciusL.Beserra No probem. – Nameless Dec 20 '12 at 18:26
How do you do to invert the expression? – Vinicius L. Beserra Dec 20 '12 at 19:20

score 3 · Answer 2 · answered Dec 20 '12 at 22:42

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

$$ -1 \times \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{x -1} = -1 \times \frac{2}{0^{+}} = -1 \times +\infty = -\infty $$

Great job. These last steps are the ones that i don´t konw. Thanks a lot. — Vinicius L. Beserra, Dec 21 '12 at 08:44

score 1 · Answer 3 · answered Dec 20 '12 at 14:06

1

As $x \to 1^{+}$, $1-x \to 0^{-}$. Hence the limit is $1/0^{-}=-\infty$.

answered Dec 20 '12 at 14:06

Siminore

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Why is the answer -$\infty$?

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