# How to compute $\lim\limits_{x \to +\infty} \left(\frac{1}{\ln (x^4) \sin (-\frac{1}{x})}\right)$?

I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?

$$\lim\limits_{x \to +\infty} \left(\frac{1}{\ln (x^4) \sin (-\frac{1}{x})}\right)$$

• what do you mean by $\sin x(- \frac{1}{x})$ ? – SiXUlm Dec 29 '15 at 12:23
• is it $\ln(x^4)$ or $(\ln(x))^4$? – Dr. Sonnhard Graubner Dec 29 '15 at 12:23
• i edited the limit – user12 Dec 29 '15 at 12:24
• Did you try l'Hopital rule? – Alex Silva Dec 29 '15 at 12:26
• Did you try $x=\frac 1y$ ? – Claude Leibovici Dec 29 '15 at 12:27

$$\lim\limits_{x \to +\infty} \left(\frac{1}{\ln (x^4) \sin (-\frac{1}{x})}\right)$$

Hint: $$=\lim\limits_{x \to +\infty} \left(\frac{1\bigg/\sin (-\frac{1}{x})}{\ln (x^4)}\right)$$ Now use L'hospital

Let $x=\frac{1}{t}\implies t\to 0^+$ as $x\to +\infty$ $$\lim_{x\to +\infty}\left(\frac{1}{\ln(x^4)\sin\left(-\frac{1}{x}\right)}\right)=\lim_{t\to 0^+}\left(\frac{1}{\ln\left(\frac 1t\right)^4\sin\left(-t\right)}\right)$$ $$=\lim_{t\to 0^+}\left(\frac{1}{4\ln(t)\sin\left(t\right)}\right)$$ $$=\frac 14\lim_{t\to 0^+}\left(\frac{t}{t\ln(t)\sin (t)}\right)$$ $$=\frac 14\lim_{t\to 0^+}\left(\frac{1}{t\ln(t)}\right)\cdot \lim_{t\to 0^+}\left(\frac{t}{\sin t}\right)$$ $$=\frac{1}{4}(-\infty)(1)=\color{red}{-\infty}$$

By L'Hopital's role we have, $$\frac{1}{4}\lim_{x\rightarrow\infty}\Big(\frac{1/(\ln(x))}{\sin(-1/x)}\Big)=\frac{1}{4}\lim_{x\rightarrow\infty}\Big(\frac{-x}{(\ln(x))^{2}\cos(-1/x)}\Big)=\frac{1}{4}\lim_{x\rightarrow\infty}\Big(\frac{-x}{(\ln(x))^{2}}\Big)=\\ \frac{1}{4}\lim_{x\rightarrow\infty}\Big(\frac{-1}{2(\ln(x))(1/x)}\Big)=\frac{1}{8}\lim_{x\rightarrow\infty}\Big(\frac{-x}{(\ln(x))^{2}}\Big)=\frac{1}{16}\lim_{x\rightarrow\infty}\Big(\frac{-1}{(\ln(x))(1/x)}\Big)=-\infty$$

Recall that $\sin t \sim_0 t$. Therefore $$L = \lim_{x \to +\infty} -\frac{x}{4\ln x} = -\infty$$ since $\ln x = o(x)$.

• This proof also works with $(\ln x)^4$. – GEdgar Dec 29 '15 at 13:06
• @GEdgar Good point, since $\ln^\alpha x = o(x)$ still holds for $\alpha \in \mathbb R$. – rubik Dec 29 '15 at 13:11