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I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?

$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$

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    $\begingroup$ Multiply top and bot by $\sqrt{x^2 - x} + \sqrt{x^2 - 1}$. Simplify. Lather, rinse, repeat. $\endgroup$ – John Hughes Dec 30 '15 at 15:07
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    $\begingroup$ The limit will tend to $-\infty$,not $+\infty$ as $x$ tends to $-\infty$ $\endgroup$ – Vinod Kumar Punia Dec 31 '15 at 9:37
  • $\begingroup$ Can someone give me, please, the link where it is asked about the limit of $(4^x-3^x)^{1/(x-1)}$ when $x$ tends to $1$?. Thank you. $\endgroup$ – Piquito Dec 31 '15 at 11:44
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$$x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)=\frac{x\left(\left(\sqrt{x^2-x}\right)^2-\left(\sqrt{x^2-1}\right)^2\right)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$

$$=\frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$

Since we're searching for the limit as $x\to -\infty$, let $x<0$. Then:

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

Because $\sqrt{1-\frac{1}{x}}\stackrel{x\to -\infty}\to 1$ and $\sqrt{1-\frac{1}{x^2}}\stackrel{x\to -\infty}\to 1$ and $x-1\stackrel{x\to -\infty}\to -\infty$.

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  • $\begingroup$ Are you "allowed" to do that? You arbitrarily set $x$ to be less than $0$ and moved on? $\endgroup$ – Nikos Feb 13 '17 at 21:12
  • $\begingroup$ @RestlessC0bra It's common sense. As $x\to -\infty$, $x$ eventually becomes always negative. $\endgroup$ – user236182 Feb 15 '17 at 0:43
  • $\begingroup$ I understand what you mean. But i have gaps of knowledge in math. As far as i'm concerned $x$ is a variable. We don't know it's actual value. It can be larger than $0$, or larger than $100$, or whatever, can it not? In this case, you say "let $x<0$". I understand why you did that. I ask, is it safe to neglect, other possible values of $x$? I'm trying to understand the math logic better. It's not common sense for me. You could say, i'm afraid to arbitrarily do things in math, that may seem logical but are mathematically wrong. If there's anything more to add, that may help me a bit, please do. $\endgroup$ – Nikos Feb 15 '17 at 9:22
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enter image description here

This is the graph of $x(\sqrt{x^2-x}-\sqrt{x^2-1})$.It is tending to $-\infty$ as $x$ is tending to $-\infty$
After rationalising
$L=\lim_{x\to -\infty}\frac{x(1-x)}{\sqrt{x^2}(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{x(1-x)}{|x|(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}$
$L=\lim_{x\to -\infty}\frac{x(1-x)}{-x(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{(1-x)}{-(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=-\infty$
Because when $x$ is negative ,$|x|=-x$,not $x$

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    $\begingroup$ I drew this graph on desmos.com/calculator $\endgroup$ – Vinod Kumar Punia Dec 31 '15 at 9:50
  • $\begingroup$ Grateful as I am about pointing out absolute value, let me share another useful grap drawing tool: rechner online: draw function graph, allowing three superimposed graphs easily $\endgroup$ – Laurent Duval Dec 31 '15 at 10:05
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The first step, always, is to get some sound intuition about the limits. You can draw a graph, but it is machine stuff and can be misleading. You'd better use tools you know, like Taylor series. Here, issues revolve around square rots. Square rots are complicated, except around $1$, where they can be linearized. So, factorize, to get square roots around $1$. You get:

$$ x |x| \left(\sqrt{1-\frac{1}{x}} - \sqrt{1-\frac{1}{x^2}} \right)\,.$$

Now, you start suspecting that the behavior could be of second order under the roots. Using $\sqrt{1-u} \sim 1-\frac{1}{2}u+\frac{1}{8}u^2+O(u^3)$, you can see that $-\frac{1}{x}$ is going to win:

$$\left(\sqrt{1-\frac{1}{x}} - \sqrt{1-\frac{1}{x^2}} \right) \sim -\frac{1}{2x}+\frac{1}{8x^2}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\right)\,.$$

Your function is likely to behave, close to $\pm\infty$ as $-\frac{|x|}{2}$ (you thus kill two birds with one stone), hence your limit will be $-\infty$.

Now you can use many different tools to continue trainin your mathematical skill, like de L'Hospital rule. One idea: divide by $-\frac{x}{2}$, and prove the product tends to $1$. The more different proofs you find, the more confident you will become in addressing novel problems.

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$$\lim_{x\to -\infty}x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)$$ $$=\lim_{x\to -\infty}x\frac{\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}$$ $$=\lim_{x\to -\infty}x\frac{\left(1-x\right)}{|x|\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$ $$=\lim_{x\to -\infty}\frac{x\left(1-x\right)}{(-x)\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$ $$=-\lim_{x\to -\infty}x\frac{\left(\frac 1x-1\right)}{\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$ $$=-\lim_{x\to +\infty}x\frac{\left(1+\frac 1x\right)}{\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}\longrightarrow \color{red}{-\infty}$$

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  • $\begingroup$ Could someone please explain the reason for what is wrong in the answer? $\endgroup$ – Harish Chandra Rajpoot Dec 31 '15 at 8:27
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    $\begingroup$ No clue... seems fine; +1 $\endgroup$ – Kugelblitz Dec 31 '15 at 9:04
  • $\begingroup$ @Kugelblitz It's because the answer is wrong. The limit is $-\infty$. See WolframAlpha. $\endgroup$ – user236182 Dec 31 '15 at 9:51
  • $\begingroup$ Oh! My phone (website and not the app) only showed me half the answer; I believed he'd stopped after a certain step as he didn't wish to give the complete solution (i.e. I believed he wished to give a hint to proceed). @user236182 [Showed until 2 steps before the last line to be precise] $\endgroup$ – Kugelblitz Dec 31 '15 at 9:57
  • $\begingroup$ Hmm.. Yes. Then it's indeed incorrect (for I upvoted the accepted answer before this answer actually). $\endgroup$ – Kugelblitz Dec 31 '15 at 9:58
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You can do the substitution $x=-1/t$ that transforms the limit into \begin{align} \lim_{x \to -\infty} x\bigl(\sqrt{x^2-x}-\sqrt{x^2-1}\,\bigr) &= \lim_{t\to0^+}-\frac{1}{t}\left(\sqrt{\frac{1}{t^2}+\frac{1}{t}}-\sqrt{\frac{1}{t^2}-1}\,\right)\\[6px] &= \lim_{t\to0^+}\frac{\sqrt{1-t^2}-\sqrt{1+t}}{t^2}\\[6px] &=\lim_{t\to0^+}\frac{1+o(t)-1-\frac{1}{2}t+o(t)}{t}\cdot\frac{1}{t}\\[6px] &=-\infty \end{align}


Comments

The substitution $x=-1/t$ will bring the limit into fraction form, usually better manageable than products like your initial form. Why not doing $x=1/t$? Because having a “positive $t$” helps in avoiding mistakes when pulling $t$ outside the square root.

You can avoid the Taylor expansion part by noticing that $$ \lim_{t\to0^+}\frac{\sqrt{1-t^2}-\sqrt{1+t}}{t} $$ is the derivative at $0$ of $$ f(t)=\sqrt{1-t^2}-\sqrt{1+t} $$ and $$ f'(t)=-\frac{t}{\sqrt{1-t^2}}-\frac{1}{2\sqrt{1+t}} $$ so $f'(0)=-\frac{1}{2}$.

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  • $\begingroup$ @user236182 I omit all terms with exponent $>1$, so $o(t)$ is good. $\endgroup$ – egreg Dec 31 '15 at 12:04
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$$\lim _{ x\to -\infty } \left( \frac { x\left( \sqrt { x^{ 2 }-x } -\sqrt { x^{ 2 }-1 } \right) \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\\ =\lim _{ x\to -\infty } \left( \frac { x\left( x^{ 2 }-x-x^{ 2 }+1 \right) }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\lim _{ x\to -\infty } \left( \frac { x\left( 1-x \right) }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\\ \\ =\lim _{ x\to -\infty } \frac { x\left( 1-x \right) }{ \left| x \right| \left( \sqrt { 1-\frac { 1 }{ x } } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } } } \right) } =\\ \lim _{ x\to -\infty } \frac { x-1 }{ \left( \sqrt { 1-\frac { 1 }{ x } } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } } } \right) } =-\infty $$

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  • $\begingroup$ The limit is $-\infty$. See WolframAlpha. $\endgroup$ – user236182 Dec 31 '15 at 9:52
  • $\begingroup$ i editted thanks for showing mistake $\endgroup$ – haqnatural Dec 31 '15 at 10:03
  • $\begingroup$ You're missing parantheses on $\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)$. $\endgroup$ – user236182 Dec 31 '15 at 10:03
  • $\begingroup$ It's still wrong. You've only edited $+\infty$ to $-\infty$, but left the other things unchanged. $\endgroup$ – user236182 Dec 31 '15 at 10:04
  • $\begingroup$ The problem is that if $x<0$, then $\sqrt{x^2}=|x|=-x$. $\endgroup$ – user236182 Dec 31 '15 at 10:08
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Given $$\lim_{x\to-\infty}(x(\sqrt{x^2-x}-\sqrt{x^2-1}))$$ As $$x(\sqrt{x^2-x}-\sqrt{x^2-1}) = \frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$ $$=\frac{x-x^2}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$ $$=\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}$$ $$\lim_{x\to-\infty}\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}=-\infty$$ As $x\to-\infty$ then $\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}\to0$ And $\frac{-1}{0}=-\infty$

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  • $\begingroup$ Worth noting the denominator is positive. $\frac{-1}{0^+}=-\infty$. There should be $^+$ on the $0$. $\endgroup$ – user236182 Dec 31 '15 at 11:09
  • $\begingroup$ This lacks rigor. You also need to let $x<1$ in order to have $\frac{1}{x^2}>\frac{1}{x^3}$ and $\frac{1}{x^2}>\frac{1}{x^4}$. $\endgroup$ – user236182 Dec 31 '15 at 11:13
  • $\begingroup$ I mean $x<-1$.${}$ $\endgroup$ – user236182 Dec 31 '15 at 16:36
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Hint: multiply by $$\frac{\sqrt{x^2-x}+\sqrt{x^2-1}}{\sqrt{x^2-x}+\sqrt{x^2-1}}.$$

You will then probably want to factor an $x^2$ out of the roots in the denominator (note that $\sqrt{x^2} = |x| = -x$ since $x<0$), so you can rewrite the above fraction as $$\frac{\sqrt{x^2-x}+\sqrt{x^2-1}}{-x\left(\sqrt{1-\frac1x}+\sqrt{1-\frac1{x^2}}\right)}.$$

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