Proving $\lim_{x \to \infty}\frac{\ln x}{x^r}=0$ and $\lim_{x \to 0^+}x^r\ln x=0$ for $r>0$ This is the question I'm trying to answer

This is how I went about proving i) and ii):
i)  $$\lim_{x \to \infty}\frac{\ln x}{x^r}=\lim_{x \to \infty}(\frac{1}{x^{r-1}}\cdot\frac{\ln x}{x})=\lim_{x \to \infty}(\frac{1}{x^{-1}}\cdot\frac{1}{x^{r}}\cdot\frac{\ln x}{x})=\lim_{x \to \infty}x\cdot\lim_{x \to \infty}\frac{1}{x^r}\cdot \lim_{x \to \infty}\frac{\ln x}{x}={\infty}\cdot\frac{1}{\infty}\cdot 0=1\cdot 0=0$$
ii) $$\lim_{x \to 0^+}x^r\ln x=\lim_{x \to 0^+}(x^{r-1}\cdot x \ln x)=\lim_{x \to 0^+}x^{r-1}\cdot \lim_{x \to 0^+}x\ln x=\lim_{x \to \infty}\frac{1}{x^{r-1}}\cdot \lim_{x \to 0^+}x\ln x=\lim_{x \to \infty}\frac{1}{x^{-1}}\cdot \lim_{x \to \infty}\frac{1}{x^r}\cdot\lim_{x \to 0^+}x\ln x=\lim_{x \to \infty}x\cdot\lim_{x \to \infty}\frac{1}{x^r}\cdot\lim_{x \to 0^+}x\ln x=\infty\cdot\frac{1}{\infty}\cdot 0=1\cdot 0=0$$
For i) I wasn't sure if writing $\lim_{x \to \infty}x$ as $\infty$ and $\lim_{x \to \infty}\frac{1}{x^r}$ as $\frac{1}{\infty}$ so that they cancelled out to give $1$ was the right thing to do since the limits for $x$ and $x^r$ are different as $x$ gets very large, although they both tend to infinity.
For ii) I wasn't sure if writing $\lim_{x \to 0^+}x^{r-1}$ as $\lim_{x \to \infty}\frac{1}{x^{r-1}}$ was right to do even though they both give the value $0$.

Using $\infty\cdot 0$ in limit formulae is not valid. Rather, perhaps, you need to show that, for $r>0$, there exists some constant $C_{r}>0$ such that $\log x\leq C_{r}x^{r/2}$ for $x\geq1$. Then $\left|\dfrac{\log x}{x^{r}}\right|\leq C_{r}\dfrac{1}{x^{r/2}}$, taking $x\rightarrow\infty$ finishes the job. For the second part, use the transformation $x\rightarrow1/x$ and appeal to the first result.
