Consider $$\lim_{n\to +\infty} n^2 2^n = 0$$
By the definition: $\forall \epsilon > 0$, $\exists M_{\epsilon} > 0$ such that $\forall n > M_{\epsilon}$ we have $|f(n) - \ell| < \epsilon$
Applying the definition I write $$n^2 2^n < \epsilon$$
Now to show this is not true $\forall \epsilon$, I thought about this: $n^2 \leq 2^n$ for $n\geq 4$. Since $n\to +\infty$ this should hold true no matter what. Then
$$n^2 2^n \leq n^4 \leq \epsilon$$
Now it suffices to take $\epsilon = 1/2$ to prove there is no such $n > M_{\epsilon}$, indeed
$$n^4 < \frac{1}{2}$$ But I get confused:
- For $n = 0$ the inequality actually holds.
- I cannot recover the form $n > M_{\epsilon}$ as asked in the definition
I still have many problems in understanding how those things work...
Thank you!