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Use the definition of Convergence of Sequences to prove that if $(x_n)\rightarrow 2$ then $$a)\qquad\left(\frac{2x_n-1}{3}\right)\rightarrow 1\\$$$$b)\qquad\qquad\left(\frac{1}{x_n}\right)\rightarrow\frac{1}{2}$$

Do I replace $x_n$ with $2$ for the sequences in parts $a$ and $b$? If so, then can I say (for part $a$):

Let $\epsilon>0$ then for a $N \in\mathbb{N}$ whenever $n \ge N$ it follows that $|1-1|<\epsilon$? Since $\left(\frac{2(2)-1}{3}\right) = 1$.

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    In some cases it helps to replace the variable with what it's approaching to get an idea for what the limit is, but if you want to formally show that part a converges to 1 you would have to show that $\forall \epsilon >0, \exists N \in \mathbb{N}, |\frac {2x_n -1}{3} - 1|<\epsilon$. Hint: Use the definition of convergence for $x_n$ to determine the convergence of part a – 2017-01-21
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    Write the definition of convergence of the first sequence, and after of the other sequences. For each $\epsilon>0$ and $N_\epsilon$ show that it holds for the other sequences. – 2017-01-21

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Hint: Observe \begin{align} \left| \frac{2x_n-1}{3}-1\right| = \left| \frac{2x_n-4}{3}\right| = \frac{2}{3}|x_n-2| \end{align} and \begin{align} \left|\frac{1}{x_n}-\frac{1}{2} \right| = \left| \frac{x_n-2}{2x_n}\right|<\frac{1}{2}|x_n-2|. \end{align}

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    So $(\frac{2x_n-1}{3})$ converges to 1 because $|\frac{2x_n-1}{3}-1|<\frac{3}{2}\epsilon$? – 2017-01-21
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    For an appropriate choice of $N$. – 2017-01-21
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    The last inequality only holds for $n$ large enough. – 2017-01-24
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No. Try to reason using $\epsilon_1$ and $\epsilon_2$ as well as $N_1$ and $N_2$:

Let $\epsilon_1 > 0$. Then we need to find an $N_1$ such that whenever $n\ge N_1$, it follows that $|\frac{2x_n-1}{3}-1| < \epsilon_1$. We know that there exists $N_2$ such that for every $\epsilon_2$…

Use a suitable $\epsilon_2$ and determine $N_1$ from that.

Of course this is not a complete solution, just a hint.