About the $\lim_{x \rightarrow \infty}(\ln{x}-x)$

Question

Find the $\lim_{x \rightarrow \infty}(\ln{x}-x)$.

We know that $\ln{x}=o(x)$ as ${x \rightarrow \infty}$ therefore we can guess that the limit will be $-\infty$.

Intuitively $x$ goes to infinity way faster than $\ln{x}$.

Here it is my formal proof of this:

We have that $\lim_{x \rightarrow \infty} \frac{x}{2\ln{x}}=+ \infty$ thus form the definition, $\exists a>0$ such that $x> 2\ln{x}$ forall $x>a$.

Now from this,$\forall x>a$, we deduce that $x- \ln{x} > 2 \ln{x}-\ln{x}=\ln{x} \Rightarrow \ln{x}-x < - \ln{x}$

Finally we have $\limsup_{x \rightarrow \infty} (\ln{x}-x) \leqslant - \infty$

Thus $\limsup_{x \rightarrow \infty} (\ln{x}-x)=\liminf_{x \rightarrow \infty} (\ln{x}-x)= -\infty$ .

Is my argument correct?

Thank you in advance!

Answer 1

for $x>0$,

$$\ln (x)-x=x (\frac {\ln (x)}{x}-1)$$

and $$\lim_{x\to+\infty}\frac {\ln (x)}{x}=0$$

thus

$$\lim_{+\infty}(\ln (x)-x)=-\infty $$