$$ \lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right) $$
I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.
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After you rationalise the expression, you can factor out an $x$ in $\sqrt{x^2-x+2}$ to get $x \sqrt{1 - 1/x + 2/x^2}$. Then the new square root term tends to $1$ as $x$ tends to infinity. – Toby Mak Oct 28 '19 at 10:58
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Note that
$$ \sqrt{x^2 - x + 2} = \sqrt{x^2 -x + \frac{1}{4} + 1.75} = \sqrt{(x - \frac{1}{2})^2 + 1.75}$$.
What is the limit of $ \sqrt{(x - \frac{1}{2})^2 + 1.75} - \sqrt{(x - \frac{1}{2})^2}$?
Alexander Geldhof
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$\lim _{x\rightarrow \infty }x- \sqrt{{x}^{2}-x+2}=\lim _{x\rightarrow \infty }{\frac { ( x-\sqrt {{x}^{2}-x+2}) ( x+\sqrt {{x}^{2}-x+2}) }{ x+\sqrt {{x}^{2}-x+2}}}=\lim _{x\rightarrow \infty }{\frac{x-2}{x+\sqrt {{x}^{2}-x+2}}}=\frac{1}{2}$
Lele Mabur
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how did the final expression become equal to 1/2? did you use L'hopitals rule?? – Eugene Oct 28 '19 at 11:04
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For infinite limit take the largest power in function, then it will show the answer since $\sqrt{x^2}=x$ – Lele Mabur Oct 28 '19 at 11:08
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No, factor out $x$ from the square root which will give $\frac{x-2}{x+x} = \frac{1}{2}$ as $x$ tends to infinity. – Toby Mak Oct 28 '19 at 11:08
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You could bound the square root between $\sqrt{x^2-2x+1}$ and $\sqrt{x^2}$. The Squeeze Theorem then gives the result. – ViHdzP Oct 28 '19 at 12:40
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Multiply by conjugate to get $\operatorname{lim}_{x\to \infty}\frac{x-2}{x+\sqrt{x^2-x+2}}=\operatorname{lim}_{x\to \infty}\frac{1-2/x}{1+\sqrt{1-1/x+2/x^2}}=1/2$
Kishalay Sarkar
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Yes rationalize is a good idea, indeed we obtain
$$x - \sqrt{x^2 - x +2 }=x - \sqrt{x^2 - x +2 }\cdot\frac{x + \sqrt{x^2 - x +2 }}{x + \sqrt{x^2 + x +2 }}=\frac{x^2-x^2+x-2}{x+ \sqrt{x^2 + x +2 }}$$
$$=\frac{x-2}{x+ \sqrt{x^2 + x +2 }}=\frac x x\frac{1-\frac 2x}{1+ \sqrt{1 + \frac1x +\frac2{x^2} }}\to \frac12$$
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Use Taylor's expansion at order $1$: for $x>0$ \begin{align} x-\sqrt{x^2-x+2}&=x-x\sqrt{1-\frac1x+\frac2{x^2}}=x-x\sqrt{1-\frac1x+o\Bigl(\frac1{x}\Bigr)}\\ &=x-x\Bigl(1-\frac1{2x}+o\Bigl(\frac1{x}\Bigr)\Bigr)=x-\Bigl(x-\frac12 +o(1)\Bigr) \\ &=\frac12+o(1). \end{align}
Bernard
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please see that the answer given by Riemann to this same question is faulty,please flag it so that it can be removed. – Kishalay Sarkar Oct 29 '19 at 03:27
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I don't think his answer is faulty. He/she points out that the notation $x\to\infty$ is ambiguous. Unless the context explains which one it is (domain of the function, for instance), it may as well mean $x\to+\infty$ as $x\to -\infty$. – Bernard Oct 29 '19 at 09:39
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all standard books including terence tao follows that by x tends to infinity they mean tends to positive infinity.When Riemann says in his answer that the limit does not exist,it does not make sense it is not $x\to a+$ and $x\to a.$ – Kishalay Sarkar Oct 29 '19 at 09:43
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Not in Bourbaki. You don't have a function defined on the projective line. – Bernard Oct 29 '19 at 09:45
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Riemann's answer that limit does not exist is not true from any sense.Who said that $lim_{x\to +\infty}$ and $lim_{x\to -\infty}$ both must exist and be equal for $lim_{x\to\infty}$ to exist??? – Kishalay Sarkar Oct 29 '19 at 09:47
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Function is not defined on extended line but limit $x\to \infty$ has its meaning. – Kishalay Sarkar Oct 29 '19 at 09:48
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Even if he follows a different convention,then also the statement that the limit does not exist makes no sense.He could ask to write plus infinity or minus infinity separately,but I have not seen someone following a weird convention like $x\to \infty$ will mean limit at both infinities will exist and be equal. – Kishalay Sarkar Oct 29 '19 at 09:54
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And I do not think that OP means to say that thing what Riemann is thinking. – Kishalay Sarkar Oct 29 '19 at 09:55
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Quite probably, but this answer has the merit of pointing out the ambiguity of the notation if nothing is explained. We've all learnt that one of the first qualities in mathematics is language accuracy. – Bernard Oct 29 '19 at 10:00
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don't you think it would be better if the first line of Riemann's answer were not there.It says the limit does not exist,it may arouse more confusion within the OP and the readers.He should edit it and point out at the ambiguity of the notation or ask to define a convetion,although this one is quite standard. – Kishalay Sarkar Oct 29 '19 at 10:02
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It may point that $\lim_{x\to\infty}$ is meaningless. B.t.w. I don't have Tao's book, but I think, by your sayings, that he specifies the meaning of $x\to\infty$ before using this notation – which,incidentally, shows the convention is not so universal as you think. – Bernard Oct 29 '19 at 10:07
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The solution lies in that the OP should define specifically if it is not so standard.And the answerer should question about its meaning rather than saying 'The limit does not exist!' – Kishalay Sarkar Oct 29 '19 at 10:08
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Can you add a comment below or a flag to Riemann's post pointing out this need for edit? – Kishalay Sarkar Oct 29 '19 at 10:10
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As you mentioned calculus, you may also substitute $x=\frac{1}{t}$ and consider the limit for $t\to 0^+$:
\begin{eqnarray*} x - \sqrt{x^2 - x +2 } & \stackrel{x=\frac{1}{t}}{=} & \frac{1-\sqrt{1-t+2t^2}}{t} \\ & \stackrel{L'Hosp.}{\sim} & -\frac{4t-1}{2\sqrt{1-t+2t^2}} \\ & \stackrel{t\to 0^+}{\rightarrow} & \frac{1}{2} \\ \end{eqnarray*}
Etemon
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trancelocation
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I would request you to check the answer to the same question given by the user Riemann.I think his answer is faulty,please check it and flag it. – Kishalay Sarkar Oct 29 '19 at 03:27
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The limit $\lim\limits_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)$ dones not exist! Because $$\lim _{x\to +\infty }x- \sqrt{{x}^{2}-x+2} =\lim _{x\to +\infty }{\frac { ( x-\sqrt {{x}^{2}-x+2}) (x+\sqrt {{x}^{2}-x+2}) }{ x+\sqrt {{x}^{2}-x+2}}}$$ $$=\lim_{x\to+\infty }{\frac{x-2}{x+\sqrt {{x}^{2}-x+2}}}=\frac{1}{2}.$$ But $$\lim _{x\to -\infty }\left(x- \sqrt{{x}^{2}-x+2}\right)=-\infty.$$
Riemann
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Please see the definition https://math.stackexchange.com/questions/2497123/definition-of-a-limit-at-infinity-in-which-x-to-infty – Kishalay Sarkar Oct 28 '19 at 13:33
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2@Riemann: You confuse $\lim_{x\to \infty} f(x)$ with $\lim_{x\to a} f(x)$ for $a \in \mathbb{R}$. For the latter we have $\lim_{x\to a} f(x)$ exists if and only if $\lim_{x\to a^+} f(x)$ and $\lim_{x\to a^-} f(x)$ exist and if these one-sided limits are equal. $\lim_{x\to \color{blue}{+\infty}} f(x)$ is within the realm of real functions nothing but $\lim_{x\to \color{blue}{+\infty}} f(x)$ and $\color{red}{\mbox{does not mean}}$ that you have to consider $\lim_{x\to \color{red}{-\infty}} f(x)$, as well. – trancelocation Oct 29 '19 at 04:23
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I don't agree. The function is defined for all $x$, hence, $\lim_{x\to\infty}$ is ambiguous, to say the least. – Bernard Oct 29 '19 at 09:44
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@Bernard $x \to \infty$ means strictly $+\infty$ if minus sign is not mentioned why will I consider $x\to -\infty$. – Kishalay Sarkar Oct 29 '19 at 09:45
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