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So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:

$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $

$\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}$

$\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1 $

The next exercise is $\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ I thought that I could solve this one in the same way by:

$\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} $

$\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}$

$\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 1 $

But apparently, the answer is $0$...

Why can't the second one be solved in the same way as the first? And why is the answer $0$?

  • 2
    the answer is $1$ – 2017-01-01
  • 0
    Can you show how apparently you are getting 0? because answer is ${1}$ – 2017-01-01
  • 0
    Your answer is correct – 2017-01-01
  • 0
    according to WolframAlpha and the answer sheet it isn't... – 2017-01-01
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    https://www.wolframalpha.com/input/?i=lim+x+to+infinity+1%2F(root(x%5E2%2B2x)-x) – 2017-01-01
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    Are you sure you have not read the function off wrongly in that case? http://www.wolframalpha.com/input/?i=Limit%5B1%2F(Sqrt%5Bx%5E2%2B2x%5D-x),x-%3Einfinity%5D – 2017-01-01
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    sorry missed a negative sign in the limit. – 2017-01-01
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    @John Bentin Can you please say me why this limit equals -1? His answer is not correct. – 2017-01-01
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    If $x \to -\infty$ the answer is 0. – 2017-01-01
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    Because eventually $x$ is negative $\sqrt{x^2}=-x$ and not $x$. – 2017-01-01
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    Is it $x\to+\infty$ or $x\to -\infty$? You have written something odd. – 2017-01-01
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    The first one is positive infinity the second one is negative infinity. @s.harp I don't get it. Could you explain your argument? – 2017-01-01
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    $\lim \limits_{x\to -\infty}\dfrac{\sqrt{x^2}}{x}$ is $-1$, not $+1$. – 2017-01-01
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    @zakaria square root of $x^2$ is $|x|$ – 2017-01-01
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    @BehrouzMaleki: The OP's answer was correct for the question originally posted. But changing the limit from $\infty$ to $-\infty$ (wrongly typeset) obviously changes the answer. – 2017-01-01
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    @John Bentin $$\lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}=0$$ and $$\lim_{x\to +\infty} \frac{1}{\sqrt{x^2+2x}-x}=1$$ Indeed this limit doesn't equal $-1$. – 2017-01-01
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    @BehrouzMaleki: Yes. We have no disagreement. – 2017-01-01

3 Answers 3

5

When $x\rightarrow-\infty$, you don't have $0$ as denominator, but $+\infty$, because when $x \rightarrow-\infty, \sqrt{x^2 -2x} \sim |x|$ and, as $x < 0$, $|x| = -x$, so you have $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.

3

We already have answers showing correct ways to solve the problem. What I found interesting was the implied question: where is the error in the steps taken in the body of the question, which appeared to show that the limit is $1$?

The steps in question are \begin{align} \newcommand{\?}{\stackrel{?}{=}} \lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} &\? \lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} \tag1\\ &\? \lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x} \tag2\\ &\? \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} \tag3\\ &\? 1 \tag4 \end{align}

The error is in the third step (from line $2$ to line $3$). In fact, if $x < 0,$ then $\sqrt{x^2} = -x,$ so \begin{align} \lim_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x} &= \lim_{x\to -\infty} \frac{-x\sqrt{1+\frac{2}{x}}+x}{2x} \\ &= \lim_{x\to -\infty} \frac{-\sqrt{1+\frac{2}{x}}+1}{2} \\ &\neq \lim_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} \end{align}

One way to avoid errors of this sort (other than constant vigilance when manipulating square roots of expressions over negative-valued variables) is to begin by observing that $$ \lim_{x\to -\infty} \frac{1}{\sqrt{x^2+2x} - x} = \lim_{y\to +\infty} \frac{1}{\sqrt{y^2-2y} + y}. $$ One can then work on the limit over $y,$ confident in the knowledge that $\sqrt{y^2}/y = 1.$

2

Let $-1/x=h$

$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=\dfrac{\sqrt{1-2h}}{\sqrt{h^2}}=\dfrac{\sqrt{1-2h}}{|h|}$

As $h\to0^+, h>0,|h|=+h$

$$\lim_{x\to-\infty}\dfrac1{\sqrt{x^2+2x}-x}=\lim_{h\to0^+}\dfrac h{\sqrt{1-2h}+1}=?$$

  • 0
    See http://math.stackexchange.com/questions/2056840/solving-lim-x-to-infty-sqrtx2x1-sqrt1-xx2-using-o-littl OR http://math.stackexchange.com/questions/1766582/evaluate-the-lim-x-to-infty-x-sqrtx2-2x/1766586#1766586 – 2017-01-01