There is an exercise I'm stuck on, also because I cannot figure out how it can be true. It says to prove that, given the sequence $n^r$ with $n$ natural, we have
$$\lim_{n\to +\infty} n^r = \begin{cases} +\infty & r > 0 \\ 1 & r = 0 \\ 0 & r < 0 \end{cases}$$
I managed to prove the first and the third cases, by using the definition of limit. But for the second one I'm stuck.
The form $\infty^0$ was undefined, we did not face Hopital theorems yet, and hence I thought of using $n^r = e^{r \ln(n)}$ but I am not sure if it can work.
- Question: is there a way to prove it with the limit definition, and not with just the calculus of the limit?
Also this made me stuck because there is another similar request, that is to consider the sequence $b^n$ and to prove that
$$\lim_{n\to +\infty} b^n = 1$$
when $b = 1$.
Which looks like to the previous exercise.
Is just the calculation of a limit a valid proof?
Actually, the $r$-power is taken before the limit, so that $$ \left.\lim_{n\rightarrow\infty} n^r\right|_{r=0} = \lim_{n\rightarrow\infty} n^0 = \lim_{n\rightarrow\infty} 1 = 1 $$