Show that if $(x_n)\rightarrow 2$ then $(1/x_n)\rightarrow 1/2$ Given a convergent sequence $(x_n)\rightarrow 2$, I am asked to prove $(1/x_n)\rightarrow 1/2$ without the use of the algebraic limit theorem. 
Here is what I have tried: Let $\epsilon>0$ be arbitrary. We must prove that $|1/x_n-1/2|<\epsilon$. Observe that $$|1/x_n-1/2|=\frac{|x_n-2|}{2|x_n|}$$
and we can make $|x_n-2|$ as small as we like. However, I'm not sure how to choose $N$, and what to do with the $|x_n|$ in the denominator. 
 A: Choose $N_{1}$ so that $|x_{n}-2|<1$ for all $n\geq N_{1}$. Then, $|x_{n}|>1$ for all $n\geq N_{1}$, so $1/|x_{n}|<1$.
Next, choose $N_{2}$ so that $|x_{n}-2|<2\epsilon$ for all $n\geq N_{2}$.
Then, take $N=\max\{N_{1},N_{2}\}$ and note that for $n\geq N$,
$$
\frac{|x_{n}-2|}{2|x_{n}|}<\frac{2\epsilon}{2}=\epsilon.
$$
A: Let $n_0$ be such that for $n \ge n_0$:
$|x_n -2| < 1$, or $-1+2 =1 < x_n <1+2=3$
Then 
$\dfrac{|x_n-2|}{2|x_n|} < (1/2) |x_n-2|.$
Let $\epsilon >0$ be given.
There is a $n_1$ such that for $n \ge n_1$:
$|x_n-2| < \epsilon.$
For $n \ge$ max$( n_0,n_1)$:
$|1/x_n-1/2| < (1/2)|x_n-2| <(1/2)\epsilon < \epsilon.$
A: Let $\varepsilon > 0$ be given. 
As $\left( x_n \right)_{n \in \mathbb{N} }$ converges to $2$, so there exists a natural number $N_1$ such that $$ \left\lvert x_n - 2 \right\rvert < 1 \tag{1} $$
for all $n \in \mathbb{N}$ such that $n > N_1$. 
Thus, for all $n \in \mathbb{N}$ such that $n > N_1$, we have 
$$ 
\begin{align}
\left\lvert x_n \right\rvert &= \left\lvert x_n - 2  + 2  \right\rvert \\
&\geq \left\lvert \ \left\lvert x_n - 2 \right\rvert - \lvert 2 \rvert \ \right\rvert  \qquad \mbox{ [ using a form of the triangle inequality for real numbers ] } \\
&= \left\lvert \ \lvert 2 \rvert  - \left\lvert  x_n - 2  \right\rvert \ \right\rvert  \\
&\geq \lvert 2 \rvert - \left\lvert x_n - 2 \right\rvert \qquad \mbox{ [ note this step ] } \\
&= 2 - \left\lvert x_n - 2 \right\rvert  \\
&> 2 - 1 \qquad \mbox{ [ using (1) ] } \\
&= 1, 
\end{align}
$$
and hence upon taking the reciprocals of the ends of the last chain of inequalities, we find that, for all $n \in \mathbb{N}$ such that $n > N_1$, we have
$$ 0 <  \frac{ 1 }{ \left\lvert x_n \right\rvert } < 1. \tag{2}  $$
$$
\begin{align}
 \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert &= \left\lvert \frac{ 2 - x_n  }{ 2 x_n  } \right\rvert \\
&= \frac{ \left\lvert  2 - x_n  \right\rvert  }{ \left\lvert  2 x_n \right\rvert   } \\ 
&= \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 \left\lvert   x_n \right\rvert   } \\ 
&< \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 }. \tag{3} \qquad \mbox{ [ using (2) above ] } 
\end{align}
$$
Now as the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converges to $2$, so there exists a natural number $N_2$ such that 
$$ \left\lvert x_n - 2 \right\rvert < 2\varepsilon \tag{4} $$
for all $n \in \mathbb{N}$ such that $n > N_2$. 
Finally, let $N$ be any natural number such that $N \geq \max \left\{ \ N_1, N_2 \ \right\}$. 
Then for all $n \in \mathbb{N}$ such that $n > N$, we have 
$$
\begin{align}
 \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert &< \frac{ \left\lvert  x_n - 2   \right\rvert  }{ 2 } \qquad \mbox{ [ using (3) above; note that $N \geq N_1$ and so $n > N_1$ also ] } \\
&< \frac{ 2 \varepsilon }{ 2 } \qquad \mbox{ [ using (4) above; note that $N \geq N_2$ and so $n > N_2$ also ] } \\
&= \varepsilon.
\end{align}
$$
Thus we have shown that, corresponding to any real number $\varepsilon > 0$, there exists a natural number $N$ such that 
$$  \left\lvert \frac{1}{x_n} - \frac{1}{2} \right\rvert < \varepsilon $$
for all $n \in \mathbb{N}$ such that $n > N$. Hence the sequence $\left( \frac{1}{x_n} \right)_{n \in \mathbb{N} }$ converges to $\frac{1}{2}$, whenever $\left( x_n \right)_{n \in \mathbb{N} }$ is any sequence of real numbers converging to $2$. 
Hope this helps.
