Can I assume this statement from this limit?

Question

So I have this limit: $$\lim_{x\to 0+} \frac{\ln(x)}{x-\sqrt x}$$

Is it correct to assume that: $$\lim_{x\to 0} \ln(x) = -\infty$$ And since for every $x$ in $\mathbb R $ if $ x > 1$ we have $x > \sqrt x$

The limit above is of type $\frac{-\infty}{0+}$ which is $-\infty$.

Answer 1

Is it correct to assume that: $$\lim_{x\to 0} \ln(x) = -\infty$$

Yes, but note that this is for $x \to 0^+$ (like in the original limit).

And since for every $x$ in $\mathbb R $ if $ x > 1$ we have $x > \sqrt x$

Although correct, you are not interested in what happens for $x>1$ since you are approaching $0$ (from the right), so $x$ will go below $1$ (eventually). You need to compare $x$ and $\sqrt{x}$ for $0<x<1$ instead of for $x>1$ and then you have $x < \sqrt{x}$. So

The limit above is of type $\frac{-\infty}{0^+}$ which is $-\infty$.

then becomes

$$\frac{-\infty}{0^-} \to +\infty$$

Answer 2

Maybe you could change your point of view and consider the change of variable $t=1/x$. Then $$\lim_{x\to 0+} \frac{\ln(x)}{x-\sqrt{x}}=\lim_{t\to +\infty} \frac{\ln(1/t)}{\frac{1}{t}-\frac{1}{\sqrt{t}}}=\lim_{t\to +\infty} \frac{\sqrt{t}\ln(t)}{1-\frac{1}{\sqrt{t}}}=+\infty.$$