I have a problem with calculating the following limit of function § $$\lim_{x\to 0} \frac{\ln(1 + \sqrt x)}{\sqrt[3] x}.$$
I have come to this moment $$\lim_{x \to 0} \frac{\frac{\ln(1+\sqrt{x})}{\sqrt{x}} \cdot \sqrt{x} }{\sqrt[3]{x}} $$ and I do not know what to do next ... I know that $\frac{\ln(1+\sqrt{x})}{\sqrt{x}}$ aims for $1$ but what to do with the ratio $\sqrt{x}/\sqrt[3]{x}$?
You are on the right tack. Yes, $\lim_{x\to 0} \frac{\ln(1 + \sqrt x)}{\sqrt{x}}=1$ and, therefore, it remains to consider the limit as $x\to 0$ of $$ \frac{\sqrt{x} }{\sqrt[3]{x}}=x^{\frac{1}{2}-\frac{1}{3}}.$$ Can you take it from here?