Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $-\infty$.

share|improve this question
2  
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
    
Consider "accepting" your favorite answers on questions by clicking the checkmark next to that solution. People will be more eager to spend time answering your questions if they see that your 'accept rate' is very high. –  rschwieb Dec 20 '12 at 15:10
1  
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
    
Thanks for the commentaries. –  Vinicius L. Beserra Dec 20 '12 at 18:21
    
But in anyway may you show me the entire job? –  Vinicius L. Beserra Dec 20 '12 at 18:29
add comment

3 Answers 3

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

share|improve this answer
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
add comment

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

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

share|improve this answer
    
Great job. These last steps are the ones that i don´t konw. Thanks a lot. –  Vinicius L. Beserra Dec 21 '12 at 8:44
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.