How to solve for the limit of $$\left(\frac{n^3}{2^n} + \frac{3}{4}\right)^n $$ using definitions.
I have seen a lot of methods using the natural number and logarithms, but we can only use rules of sequences and definitions.
I started by attempting to separate the inside but don’t really know what to do / where to go.
(note: we can make use of a few standard null sequences, algebra of limits, sandwich theorem)
On
We have that
$$\left(\frac{n^3}{2^n} + \frac{3}{4}\right)^n =e^{n\log\left(\frac{n^3}{2^n} + \frac{3}{4}\right)} \to 0$$
indeed since $\frac{n^3}{2^n}\to 0$
$$n\log\left(\frac{n^3}{2^n} + \frac{3}{4}\right) \to \infty\cdot \log \left(\frac{3}{4}\right)=-\infty$$
As an alternative since $\frac{n^3}{2^n}\to 0$, we have that for some $n_0$ for all $n\ge n_0 \implies \frac{n^3}{2^n}\le \frac18$ then for $n\ge n_0$
$$\left(\frac{n^3}{2^n} + \frac{3}{4}\right)^n \le\left(\frac18+\frac34\right)^n=\left(\frac78\right)^n \to 0$$
On
Finding the limit is very straight forward.
The inner function approaches $3/4$ because $$\frac {n^3}{2^n}\to 0$$
Therefore the sequence approaches $0$
Proof by definition is complicated if by difinitin you mean to show for a given $\epsilon $ finding a natural number $N$ such that $$n\ge N \implies \left(\frac{n^3}{2^n} + \frac{3}{4}\right)^n <\epsilon$$
We can approach solving it in two steps, one for the inner sequence and one for the outer one and somehow combine them.
The sequence $\left(\frac{n^3}{2^n}\right)_{n\in\mathbb N}$ is decreasing if we restrict it to $\mathbb N\setminus\{1,2,3\}$. It turns out that $\frac{15^3}{2^{15}}<\frac3{20}$ and that therefore $\frac{15^3}{2^{15}}+\frac34<\frac{18}{20}=\frac9{10}$. So,$$n\geqslant15\implies\left(\frac{n^3}{2^n}\right)^n<\left(\frac9{10}\right)^n=\frac1{\left(\frac{10}9\right)^n}=\frac1{\left(1+\frac19\right)^n}<\frac9n.$$Can you take it from here?