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I'm trying to calculate the following limit:

$$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log(1+{e}^{3\sqrt x})}$$

For WolframAlpha the result is: $\frac 13$, I've seen its step by step process but I didn't understand the background logic.

Before I got stuck, I did this step:

$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log[{e}^{3\sqrt x}({e}^{-3\sqrt x}+1)]}$ and then $\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log({e}^{3\sqrt x})+\log({e}^{-3\sqrt x}+1)}$

so $$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle {3\sqrt x}}$$

Someone could give me a hint for solve it?

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    $\begingroup$ Did you perhaps make a typo? Do you mean $x^6 + 2x^4$ in the square root? $\endgroup$
    – Marcus M
    Jul 14, 2017 at 16:34
  • $\begingroup$ @MarcusM No, it is correct. I've checked it another time. $\endgroup$
    – NapMaster
    Jul 14, 2017 at 16:38
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    $\begingroup$ it looks like the numerator has since been changed to have $-x$ rather than $+x$, which does the same thing. $\endgroup$
    – Marcus M
    Jul 14, 2017 at 16:42

4 Answers 4

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First note that $\sqrt{x^6-2x^4} = x^3 \sqrt{1 - 2/x^2} = x^3 \left(1 - 1/x^2 + O(1/x^4)\right) = x^3 - x + O(1/x)$ so $$\sqrt x + x^3 -\sqrt{x^6-2x^4} - x = \sqrt x + x^3 - \left( x^3 - x + O(1/x) \right) - x \\ = \sqrt x + O(1/x) \sim \sqrt x$$ for large $x$.

Also, for large $x$, $\log\left(1+e^{3\sqrt x}\right) \sim 3\sqrt x$.

Therefore, $$\frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} - x}{\log(1+{e}^{3\sqrt x})} \sim \frac{\sqrt x}{3\sqrt x} = \frac{1}{3}.$$

Thus the limit is $\frac13$.

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  • $\begingroup$ The original poster has changed the numerator to read as $-x$, so it actually cancels when you do that calculation. $\endgroup$
    – Marcus M
    Jul 14, 2017 at 16:51
  • $\begingroup$ Yes, sorry. Anyway the result should be $\frac 13$ $\endgroup$
    – NapMaster
    Jul 14, 2017 at 16:51
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    $\begingroup$ I've modified my answer according to the change of sign. $\endgroup$
    – md2perpe
    Jul 14, 2017 at 16:57
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You are proceeding in the right direction. Your last expression can be rewritten as $$\frac{1}{3}+\frac{\sqrt{x}(x^{2}-\sqrt{x^{4}-2x^{2}}-1)}{3}$$ Thus we need to deal with the numerator of the last term. We will show that it tends to $0$ as $x\to\infty$. To do so in an easy manner we can put $x=1/t$ so that $t\to 0^{+}$ and then the numerator is transformed into $$\frac{1-t^{2}-\sqrt{1-2t^{2}}} {t^{2}\sqrt{t}}$$ and then we need the rationalization trick to express it as $$\frac{(1-t^{2})^{2}-(1-2t^{2})}{t^{5/2}(1-t^{2}+\sqrt{1-2t^{2}})}$$ which is further simplified as $$\frac{t^{3/2}}{1-t^{2}+\sqrt {1-2t^{2}}}$$ and this tends to $0$ because numerator tends to $0$ and denominator tends to $2$.

The answer to the problem is thus $1/3$.

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  • $\begingroup$ This is easier but a bit longer. If it is not a problem, could you link me a theory page where I can see other "rationalization tricks", included this one? I just know few ones. $\endgroup$
    – NapMaster
    Jul 15, 2017 at 11:28
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    $\begingroup$ @NapMaster: Well, I don't know many tricks apart from the usual multiplication by conjugate. Most of the times there are other ways to deal with problem, for example using the limit formula $$ \lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$$ $\endgroup$
    – Paramanand Singh
    Jul 15, 2017 at 11:42
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HINT: Break it up as $$\frac{\sqrt{x}}{3\sqrt{x}} + \frac{x^3 - \sqrt{x^6 - 2x^4} - x}{3\sqrt{x}}.$$

To deal with the numerator of the second term, write it as $$x^3\left(1 - \sqrt{1 - 2/x^2} - 1/x^2\right)$$

and use the Taylor expansion of $\sqrt{1 - y}$.

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  • $\begingroup$ Is there a way to solve it without using Taylor? $\endgroup$
    – NapMaster
    Jul 14, 2017 at 16:45
  • $\begingroup$ You could use L'Hospital's rule, but they're basically the same thing. $\endgroup$
    – Marcus M
    Jul 14, 2017 at 16:47
  • $\begingroup$ Sorry, why does the denominator of the second term is $\sqrt x$ and not ${3\sqrt{x}}$? $\endgroup$
    – NapMaster
    Jul 14, 2017 at 16:55
  • $\begingroup$ @NapMaster, Oh that's just a typo. I've fixed it. $\endgroup$
    – Marcus M
    Jul 14, 2017 at 18:14
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If you multiply/divide by the conjugate $\sqrt x+x^3+\sqrt{x^6-2x^4}-x$, the numerator becomes

$$(\sqrt x+x^3-x)^2-(x^6-2x^4)=2x^{7/2}+\cdots$$

It is enough to keep the leading term.

The conjugate can be written as $$2x^3+\cdots$$ by pulling $x^3$ out of the radical.

Then with the denominator $3\sqrt x$, the limit is

$$\frac{2x^{7/2}}{2x^33\sqrt x}\to\frac13.$$

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