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I'm trying to prove the following:

Using the known limits of $\exp: \mathbb{R} \to (0,\infty)$ for $x\to \pm\infty$ and $\ln : (0,\infty) \to \mathbb{R} $ for $x\to 0^+,\infty$, show that for any $r\in \mathbb{R}$ with $r>0$ we have $$ \lim_{x\to\infty}x^r = \infty ~~~~\text{and}~~~~ \lim_{x\to0^+}x^r = 0 $$

This is how I proved both:

  1. $$ \lim_{x \rightarrow \infty}x^r=\lim_{x \rightarrow \infty}e^{rlnx}=e^{r\lim_{x \rightarrow \infty}\ln x}=e^{r\infty}=(e^r)^\infty=\infty \ \text{since $r>0$}$$

  2. $$\lim_{x \rightarrow 0^+}x^r=\lim_{x \rightarrow 0^+}e^{r\ln x}=e^{r\lim_{x \rightarrow 0^+}\ln x}=e^{r(-\infty)}=(e^r)^{-\infty}=\frac{1}{(e^r)^{\infty}}=\frac{1}{\infty}=0 \ \text{since $r>0$} .$$

I'm not sure if raising $e^r$ to positive or negative infinity is valid here.

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Your intuition is correct, however you can write the solution a little better:

  1. Since $x^r={e^{\ln(x^r)}}=e^{r\ln(x)}$ and $\displaystyle \lim_{x \to \infty} \ln(x)=+\infty$ then $\displaystyle \lim_{x \to \infty} x^r=+\infty$.

  2. Again, we have $x^r=e^{r\ln(x)}$, and since $\displaystyle \lim_{x \to 0^+} \ln(x)=-\infty$ then $\displaystyle \lim_{x \to 0^+} x^r=0$.

It's important to clarify that $\pm \infty$ ARE NOT REAL NUMBERS, so don'treat them as such (unless you are working in the extended real line $\overline{\mathbb{R}}$). Some teachers are really picky with that.

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