$$\lim_{x \to +\infty}\left(x^\frac{7}{6}-x^\frac{6}{7}\cdot \ln^2( x) \right)$$ I can not decide the limit. I understand that it is necessary to apply L'Hôpital's rule, when there will be a fraction. But to start, how to make this shot in this example?

Please help me to solve it L'Hospital's rule!

  • 2
    $\begingroup$ $ln^2 x$ : $(ln\ x)^2$ or $ln\ x^2 $ ? $\endgroup$
    – HK Lee
    Mar 21, 2015 at 15:59
  • 2
    $\begingroup$ This is $$(lnx)^2$$ $\endgroup$
    – andre1
    Mar 21, 2015 at 16:01
  • $\begingroup$ Actually I would have interpreted $\ln^2(x)$ as $\ln\ln x$. $\endgroup$
    – celtschk
    Mar 21, 2015 at 16:03
  • 1
    $\begingroup$ the limit remains the same if we replace $x$ by $x^{42}$ and you will get $x^{49}-42^2x^36ln(x)^2$ which is effectively $+\infty$ because of th comparison between powers, For the question I don't think that there is a theorem saying that :"every limit can be computed using L'Hopital's rule" $\endgroup$
    – Elaqqad
    Mar 21, 2015 at 16:10
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    $\begingroup$ Intuitively, $x^{\epsilon}$ eventually grows "faster" than $\ln^2 x$ for any $\epsilon > 0$, so the left term dominates the right term since its power is larger and the $\ln^2 x$, as $x$ gets sufficiently large, doesn't have an effect. $\endgroup$
    – MT_
    Mar 21, 2015 at 16:29

1 Answer 1


The first thing to do is a substitution: set $x=t^{42}$, so $x^{6/7}=t^{36}$, $x^{7/6}=t^{49}$ and $\ln x=42\ln t$. Then the limit becomes $$ \lim_{t\to\infty}(t^{49}-t^{36}\cdot 42^2(\ln t)^2)= \lim_{t\to\infty}t^{49}\left(1-1764\frac{(\ln t)^2}{t^{13}}\right) $$ Now we have something to which we can apply l’Hôpital: $$ \lim_{t\to\infty}\frac{(\ln t)^2}{t^{13}}= \lim_{t\to\infty}\frac{2\ln t}{13t^{13}}= \lim_{t\to\infty}\frac{2}{169t^{13}}=0 $$

  • $\begingroup$ Unfortunately, it is usually necessary to use $\endgroup$
    – andre1
    Mar 21, 2015 at 16:13
  • $\begingroup$ @andre1 I have no problem whatsoever with the late Marquis de l’Hôpital and I let students freely use his theorem, when it's helpful. $\endgroup$
    – egreg
    Mar 21, 2015 at 16:15
  • $\begingroup$ But my teacher too principled. He is required to use this rule. $\endgroup$
    – andre1
    Mar 21, 2015 at 16:17
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    $\begingroup$ Didn't you mix up the roles of $t^{49}$ and $t^{36}$ here? $\endgroup$ Mar 21, 2015 at 16:22
  • $\begingroup$ @HansLundmark Thanks, I'll fix it! $\endgroup$
    – egreg
    Mar 21, 2015 at 16:22

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