Solutions

(a) By definition,

$$\begin{aligned} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N &\implies \abs{x_{n} - 2} < \frac{3\epsilon}{2} \\ &\implies \frac{2}{3}\abs{x_{n} - 2} < \epsilon \\ &\implies \abs{\frac{2x_{n} - 4}{3}} = \abs{\frac{2x_{n} - 1}{3} - 1} < \epsilon \end{aligned}$$

(b) We can solve this using the same approach that was used to proove one of the algebraic limit theorems. That is, notice first that

$$\abs{\frac{1}{x_{n}} - \frac{1}{2}} = \abs{\frac{2 - x_{n}}{2x_{n}}} = \abs{x_{n} - 2} \frac{1}{\abs{2x_{n}}}.$$

In our attempt to prove that $\frac{1}{x_{n}} \rightarrow \frac{1}{2}$, we will need to find strict upper bounds on both terms of the product.

By definition, for any $\epsilon_1 > 0$, there exists $N_1$ for which the numerator $\abs{x_{n} - 2}$ is strictly bound by $\epsilon_1$ for all $n \geq N$.

For the denominator, since $x_{n}$ converges to 2, we can choose $\epsilon_2 = \frac{3}{2}$, in which case there exists $N_2$ where $x_{n} \in V_{\frac{3}{2}}(2)$ for all $n \geq N_2$. This would then imply an upper bound of $1$ for $\frac{1}{\abs{2x_{n}}}$.

For any $\epsilon > 0$, by choosing $N = \max \{N_1, N_2\}$, we ensure that the definition of convergence to 2 is satisfied for this sequence.