Friday, April 5, 2019

calculus - exponential limit comparison

Its straightforward to see that $lim_{n\rightarrow \infty}\frac{n^{m-1}}{n^m} = 0$, $m$ is a fixed positive integer. This meas that $n^m$ grows faster that $n^{m-1}$.



Now, let be $\{a_0,a_1,...,a_m\}$ rational numbers, consider the following limit:

$lim_{n\rightarrow \infty} a_0+a_1 n+a_2 n^2+...+a_m n^m$.
It would seems to me that this goes to $+\infty$ or $-\infty$. Since $a_m n^m$ grows (decreases) faster than any other summand, we just check the sign of $a_m $.
How can I prove this? Any suggestion?

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...