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Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$

First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.

One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.

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Well,

$$\sqrt{x^6 - 9 x^3} = x^{3} \left(1 - \frac{9}{x^{3}}\right)^{1/2}$$

and so the question is really about understanding how $\sqrt{1 - t}$ looks when $t$ is pretty small. By noticing that the derivative $\frac{d}{dt} \sqrt{t} = \frac 1 2$ when $t = 1$, we can say that

$$\left(1 - \frac{9}{x^{3/2}}\right)^{1/2} \approx \frac 1 2 \left(\frac{-9} {x^{3}}\right)$$

which leads quickly to the claimed limit of $-9/2$.


Alternatively, there is a completely standard technique of multiplying and dividing by the conjugate, which is $\sqrt{x^6 - 9x^3} + x^3$, but I find that this isn't very enlightening.

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    $\begingroup$ Or substituting $t=1/x^3$, which brings the limit in the form $\lim\limits_{t\to0^+}\dfrac{\sqrt{1-9t}-1}{t}$ which is a derivative. $\endgroup$
    – egreg
    Commented Sep 5, 2016 at 16:32
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$$\sqrt{x^6-9x^3}-x^3=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3}$$

As $x$ goes to $\infty$, $\sqrt{x^6-9x^3} \approx x^3.$

More, precisely, divide the numerator and denominator by $x^3$:

\begin{align} \sqrt{x^6-9x^3}-x^3&=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}\\&=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3} \\ & =\frac{-9}{\sqrt{1-9x^{-3}}+1} \end{align}

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Hmmm, no one has pointed out the obvious:

$\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$

$ = \lim \sqrt{(x^3 - 9/2)^2} - x^3 = \lim x^3 - 9/2 - x^3 = -9/2$

Intuition.... hmm .... I guess realizing the $x^3$ from $\sqrt {x^6 + stuff}$ was going to cancel the $-x^3$. So I want some $\lim \sqrt {Y_{x^3}^2} - x^3$ and figuring $Y_{x^3}^2$ must be.

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So why does $\lim \sqrt{x^6 -9x^3} = \lim\sqrt{x^6 -9x^3 + 36/4}$?

Let $\epsilon > 0$. We wish to solve for which $x$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| = x^3 - \frac 92 - \sqrt{x^6-9x^3} < \epsilon$. If we can show that this can be solved for all $x > M$ for some $M$ we are done.

$(x^3 - \frac 92) - \epsilon < \sqrt{x^6 - 9x^3} $. For large enough $x$ we may assume this are both positive.

$x^6 - 9x^3 + 9 - 2\epsilon*(x^3 - \frac 92) + \epsilon^2 < x^6 - 9x^3$

$9 - 2\epsilon*(x^3 - \frac 92) + \epsilon^2 < 0$

$(x^3- \frac 92) > \frac{(9 + \epsilon^2)}{2\epsilon}$

So for any $x > \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ this will be true.

So for all $\epsilon > 0$ if $x > M = \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| < \epsilon$

So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}= 0$

So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}= \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}$

So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}-x^3 = \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4} -x^3 = -\frac 92$

Or more generally...

If $c_x \rightarrow \infty$ and $f$ is continuous, then $\lim_{x\rightarrow \infty}f(c_x + h) = \lim_{x\rightarrow \infty}f(c_x(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x(\lim_{x_\rightarrow \infty}(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x*1)=\lim_{x\rightarrow \infty}f(c_x)$

Let $c_x = x^6 - 9x$, $h= 9=36/4$ $f(y) = \sqrt{y}$.

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  • $\begingroup$ It seems somewhat obvious, but how would you prove $\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$? $\endgroup$
    – D. W.
    Commented Sep 5, 2016 at 6:27
  • $\begingroup$ It's intuitive lim f(x+h) = f(x) as x goes to infinity for cont. because the h becomes insignificant and I think it's proven somewhere. $\endgroup$
    – fleablood
    Commented Sep 5, 2016 at 6:59
  • $\begingroup$ let f(x) -> c let e > 0 that for x > M, |f(x) - c| < e. Then x + h > x > M so |f(x+h) - c| < e so f(x + h) -> c. That's quick and dirty. Can do the same if f(x) -> infinity. $\endgroup$
    – fleablood
    Commented Sep 5, 2016 at 7:11
  • $\begingroup$ could you specify what $f$ and $h$ one should take in your lemma to prove the equality $\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$? $\endgroup$
    – timon92
    Commented Sep 5, 2016 at 7:25
  • $\begingroup$ Actually I did that wrong. $\endgroup$
    – fleablood
    Commented Sep 5, 2016 at 15:57
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Its best to have a look at this answer to a similar question. Apart from that I want to add that most of the stuff in calculus/real-analysis is not to be handled by intuition. Even mathematicians have been led astray with their intuition in real-analysis (there exist continuous everywhere but differentiable nowhere functions, there exist differentiable functions whose derivatives are not integrable in sense of Riemann).

In particular the concept of limit is also not as intuitive as one might think of (i.e. on a level of $+, -, \times, /$). A limit is always evaluated using theorems which deal with evaluation of limits and nothing else. And worst approach to limits is to think of them as an approximation and hence $\sqrt{x^{6} - 9x^{3}}$ can't be replaced by $x^{3}$ because they are approximately equal. The simplest and straight-forward approach to such simple limit questions is the use of standard limits. Here we use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ First we replace $x^{3}$ by $1/h$ and as $x \to \infty$ we have $h \to 0^{+}$. Then \begin{align} L &= \lim_{x \to \infty}\sqrt{x^{6} - 9x^{3}} - x^{3}\notag\\ &= \lim_{h \to 0^{+}}\sqrt{\frac{1}{h^{2}} - \frac{9}{h}} - \frac{1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{\sqrt{1 - 9h} - 1}{h}\notag\\ &= \lim_{t \to 1^{-}}\dfrac{t^{1/2} - 1}{\dfrac{1 - t}{9}}\text{ (putting }1 - 9h = t)\notag\\ &= -9\lim_{t \to 1^{-}}\frac{t^{1/2} - 1}{t - 1}\notag\\ &= -9 \cdot\frac{1}{2}\text{ (using (1))}\notag\\ &= -\frac{9}{2}\notag \end{align}

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The only thing I would add to T. Bongers' answer is that the key to these kinds of questions is generally to ask:

  1. How can I transform the limit to evaluate it using standard techniques?
  2. Especially with limits at infinity, what is 'dying off' quicker?
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For a fixed positive $k$ we should expect $\sqrt {y+k}$ to be very close to $\sqrt y$ when $y$ is very large, because when $d$ is not close to $0,$ we expect a big difference between $(d+\sqrt y)^2$ and $y,$ which means $d+\sqrt y$ is too big to be $\sqrt {y+k}$.

We can confirm this by observing that if $k,y$ are positive and $\sqrt {y+k}=\sqrt y+e_y$ then $$y+k=(\sqrt y +e_y)^2=y +2e_y\sqrt y +e_y^2>y+2e_y\sqrt y$$ which implies $0<e_y<k/2\sqrt y.$

With $y=x^6-9x^3$ and $k=(9/2)^3,$ we have $x^3-9/2=\sqrt {y+k}=e_y+\sqrt y.$

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For intuition, first get rid of the annoying cubes,

$$\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3=\lim_{x\to\infty}\sqrt{x^2 - 9x}-x.$$

Then the classical trick is to multiply/divide by the conjugate binomial and simplify,

$$\lim_{x\to\infty}\left(\sqrt{x^2 - 9x}-x\right)\frac{\sqrt{x^2 - 9x}+x}{\sqrt{x^2 - 9x}+x}=\lim_{x\to\infty}\frac{-9x}{\sqrt{x^2 - 9x}+x}=-\frac92.$$

Another approach is to factor out $x$ and use the Taylor development $\sqrt{1+t}\approx 1+t/2$:

$$\lim_{x\to\infty}x\left(\sqrt{1 - \frac9x}-1\right)=\lim_{x\to\infty}x\left(- \frac9{2x}\right).$$


You can "feel" why the answer is not $0$ by comparing the plots of the functions $\sqrt{x^2-9x}$ and $x$.

enter image description here

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  • $\begingroup$ Why this downvote ?? $\endgroup$
    – user65203
    Commented Sep 5, 2016 at 17:46
  • $\begingroup$ I did not issue the -1, but I would use a different variable name like $y$ for $x^3$ and explicitly define $y=x^3$. This syntax (and that's the only real problem, syntax) is preferred over reusing a variable name as you did. $\endgroup$ Commented Sep 5, 2016 at 17:52
  • $\begingroup$ @OscarLanzi From a didactical point of view I can agree. From a formal point of view, there's nothing wrong reusing the same dummy variable (I actually did it on purpose). Thanks for the comment. $\endgroup$
    – user65203
    Commented Sep 5, 2016 at 18:29
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T. Bongers' answer is right, but a bit sloppy. A better way would be to use the binomial expansion.

$$\sqrt{1-9x^{-3}} = \sum_{n=0}^\infty \frac{(1/2)^{\color{red}{n}}(-9)^nx^{-3n}}{n!} = 1 - \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots$$ Where the red thing represents a falling power. $$\lim_{x\to+\infty}\sqrt{x^6-9x^3}-x^3=\lim_{x\to+\infty}x^3(\sqrt{1-9x^{-3}} - 1) = \lim_{x\to+\infty}x^3\left(- \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots\right) = -\frac{9}{2}$$

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    $\begingroup$ FWIW, I object to "a bit sloppy" and to "better way". Re the approach you suggest, the full expansion in particular, seems quite irrelevant. $\endgroup$
    – Did
    Commented Sep 5, 2016 at 5:44

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