1
$\begingroup$

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?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ Did you perhaps make a typo? Do you mean $x^6 + 2x^4$ in the square root? $\endgroup$ – Marcus M Jul 14 '17 at 16:34
  • $\begingroup$ @MarcusM No, it is correct. I've checked it another time. $\endgroup$ – NapMaster Jul 14 '17 at 16:38
  • 1
    $\begingroup$ 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 '17 at 16:42
2
$\begingroup$

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 $\frac13$.

Improve this answer
$\endgroup$
  • $\begingroup$ 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 '17 at 16:51
  • $\begingroup$ Yes, sorry. Anyway the result should be $\frac 13$ $\endgroup$ – NapMaster Jul 14 '17 at 16:51
  • 1
    $\begingroup$ I've modified my answer according to the change of sign. $\endgroup$ – md2perpe Jul 14 '17 at 16:57
1
$\begingroup$

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 $1/3$.

Improve this answer
$\endgroup$
  • $\begingroup$ 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 '17 at 11:28
  • 1
    $\begingroup$ @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 '17 at 11:42
0
$\begingroup$

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 $\sqrt{1 - y}$.

Improve this answer
$\endgroup$
  • $\begingroup$ Is there a way to solve it without using Taylor? $\endgroup$ – NapMaster Jul 14 '17 at 16:45
  • $\begingroup$ You could use L'Hospital's rule, but they're basically the same thing. $\endgroup$ – Marcus M Jul 14 '17 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 '17 at 16:55
  • $\begingroup$ @NapMaster, Oh that's just a typo. I've fixed it. $\endgroup$ – Marcus M Jul 14 '17 at 18:14
0
$\begingroup$

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

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.