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 $0$.

Your question has already given a big hint. You may assume that these results hold for $r=1$. Thus we are given the following limits $$\lim_{x\to\infty} \frac{\log x} {x} =0=\lim_{x\to 0^{+}}x\log x$$ With these as given the problem is too simple. We have $$\lim_{x\to \infty} \frac{\log x} {x^{r}} =\frac{1}{r}\lim_{x\to\infty}\frac{\log x^{r}} {x^{r}} =\frac{1}{r}\lim_{t\to\infty}\frac{\log t} {t} =\frac{1}{r}\cdot 0=0$$ We have used the substitution $t=x^{r}$ and it is given that $t\to\infty$ as $x\to\infty$. Similarly the second limit is evaluated. 
