4
$\begingroup$

$$\lim_{x \to +\infty}\left(x^\frac{7}{6}-x^\frac{6}{7}\cdot \ln^2( x) \right)$$ 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!

$\endgroup$
  • 2
    $\begingroup$ $ln^2 x$ : $(ln\ x)^2$ or $ln\ x^2 $ ? $\endgroup$ – HK Lee Mar 21 '15 at 15:59
  • 2
    $\begingroup$ This is $$(lnx)^2$$ $\endgroup$ – andre1 Mar 21 '15 at 16:01
  • $\begingroup$ Actually I would have interpreted $\ln^2(x)$ as $\ln\ln x$. $\endgroup$ – celtschk Mar 21 '15 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 '15 at 16:10
  • 1
    $\begingroup$ 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. $\endgroup$ – MCT Mar 21 '15 at 16:29
3
$\begingroup$

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: $$ \lim_{t\to\infty}\frac{(\ln t)^2}{t^{13}}= \lim_{t\to\infty}\frac{2\ln t}{13t^{13}}= \lim_{t\to\infty}\frac{2}{169t^{13}}=0 $$

$\endgroup$
  • $\begingroup$ Unfortunately, it is usually necessary to use $\endgroup$ – andre1 Mar 21 '15 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 '15 at 16:15
  • $\begingroup$ But my teacher too principled. He is required to use this rule. $\endgroup$ – andre1 Mar 21 '15 at 16:17
  • 2
    $\begingroup$ Didn't you mix up the roles of $t^{49}$ and $t^{36}$ here? $\endgroup$ – Hans Lundmark Mar 21 '15 at 16:22
  • $\begingroup$ @HansLundmark Thanks, I'll fix it! $\endgroup$ – egreg Mar 21 '15 at 16:22

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.