Prove the limit of a piecewise defined sequence Is my approach correct?
Define the sequence $\{b_{n}\}$ as
$$ b_n =\begin{cases} \frac{n^4 + 1}{n^4}&  \text{if $n$ is even}\\
    \frac{n^2 - 1}{n^2}        &   \text{if $n$ is odd}\end{cases}.$$
Prove that $\lim b_n = 1$
Proof. We partition $\{b_{n}\}$ into subsequences $\{b_{2n}\}$ and $\{b_{2n-1}\}$ where
$$\{b_{2n}\} = \{\frac{n^4 + 1}{n^4}\}$$
$$\{b_{2n-1}\} = \{\frac{n^2- 1}{n^2}\}.$$
Taking the limits,
$$\lim b_{2n} = \lim \frac{n^4 + 1}{n^4} = \lim 1 + \frac{1}{n^4} = 1$$
$$\lim b_{2n-1} = \lim \frac{n^2 - 1}{n^2} = \lim 1 - \frac{1}{n^2} = 1.$$
Let $\varepsilon > 0$. Then there exists $N_{1},N_{2} \in \mathbb{N}$ s.t.
$$|b_{2n} - 1| < \varepsilon,\forall 2n \geq N_{1}$$
$$|b_{2n-1} - 1| < \varepsilon,\forall 2n-1 \geq N_{2}.$$
Let $N = \max\{N_{1},N_{2}\}$ and $n \geq N.$
If $n$ is even, then $n = 2j$ for some $j \in \mathbb{N}$ and
$$|b_{n} - 1| = |b_{2j} - 1| < \varepsilon.$$
If $n$ is odd, then $n = 2j - 1$ for some $j \in \mathbb{N}$ and
$$|b_{n} - 1| = |b_{2j - 1} - 1| < \varepsilon.$$
From both cases, $|b_{n} - 1| < \varepsilon. \square$

Yes, I do not see any particular mistake. The key point is indeed to notice the two subsequences may have different rates of convergence and hence if you take $N$ the maximum of $N_1$ and $N_2$, then you are sure you get the definitions of convergence in both cases, hence for all terms greater then $N$.
