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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 $0$?

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    $\begingroup$ the answer is $1$ $\endgroup$ – ThePortakal Jan 1 '17 at 14:11
  • $\begingroup$ Can you show how apparently you are getting 0? because answer is ${1}$ $\endgroup$ – Fawad Jan 1 '17 at 14:14
  • $\begingroup$ Your answer is correct $\endgroup$ – John Bentin Jan 1 '17 at 14:15
  • $\begingroup$ according to WolframAlpha and the answer sheet it isn't... $\endgroup$ – Zakaria el Ninja Jan 1 '17 at 14:15
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    $\begingroup$ sorry missed a negative sign in the limit. $\endgroup$ – Zakaria el Ninja Jan 1 '17 at 14:17
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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 $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.

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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 $\sqrt{y^2}/y = 1.$

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

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

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