To evaluate $\lim_{x \to 0^+} \frac{\log(x)}{\sqrt x}$ using inequality

Question

To evaluate $$\lim_{x \to 0^+} \frac{\log(x)}{\sqrt x}$$

I know that $\log(x) < x$ for $x > 0$. So dividing by square root of $x$ and taking limits gives me nothing. Which inequality should I use here?

Thanks

Answer 1

If we substitute $t=\frac{1}{x}$ then $$\lim_{x \to 0^+} \frac{\log(x)}{\sqrt x}=\lim_{t \to +\infty} {(-\sqrt{t}\cdot\log{t})} = -\infty$$ since $\log{t}>1$ for $t>e.$

Answer 2

$$\lim_{x \to 0^+} \log(x)=-\infty$$ $$\lim_{x \to 0^+} \sqrt{x}=0^+$$ $$\lim_{x \to 0^+} \frac {\log(x)}{\sqrt{x}} =-\infty$$ You don't need inequality here