# Limit $\lim_{x \to +\infty}\left(x^\frac{7}{6}-x^\frac{6}{7}\cdot \ln^2( x) \right)$ using L'Hôpital's rule.

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

• $ln^2 x$ : $(ln\ x)^2$ or $ln\ x^2$ ? – HK Lee Mar 21 '15 at 15:59
• This is $$(lnx)^2$$ – andre1 Mar 21 '15 at 16:01
• Actually I would have interpreted $\ln^2(x)$ as $\ln\ln x$. – celtschk Mar 21 '15 at 16:03
• 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" – Elaqqad Mar 21 '15 at 16:10
• 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. – MCT Mar 21 '15 at 16:29

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: $$\lim_{t\to\infty}\frac{(\ln t)^2}{t^{13}}= \lim_{t\to\infty}\frac{2\ln t}{13t^{13}}= \lim_{t\to\infty}\frac{2}{169t^{13}}=0$$
• Didn't you mix up the roles of $t^{49}$ and $t^{36}$ here? – Hans Lundmark Mar 21 '15 at 16:22