- Thread starter
- #1
How do I now show $a > 1$?To prove this limit, we need to show that exists $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have that
$$\left| \frac{1}{a^n} - 0 \right| < \varepsilon,$$
for all $\varepsilon >0$.
Not sure what tools you have available, but if perhaps you could do
$$\frac{1}{a^n} < \varepsilon \leadsto a^n > \frac{1}{\varepsilon} \leadsto n \log_a a = n > \log_a \left( \frac{1}{\varepsilon} \right).$$
Therefore, take $N_0 = \left\lceil \log_a \left( \frac{1}{\varepsilon} \right) \right\rceil$.
Not entirely sure, but the whole process looks okay.
Since you said "if $a>1$", it seems like you're given this information. It is your hypothesis. It is because of this that we can take $\log_a r$.$\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$ if $a > 1$.
Not sure how to handle this one. Do I want have $\frac{1}{\sqrt[n]{\epsilon}} < a$?