This is the question I'm trying to answer
This is how I went about proving i) and ii):
i) lim
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.
Answer
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.
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