Show that if $(x_n)\rightarrow 2$ then $(1/x_n)\rightarrow 1/2$

Question

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.

Answer 1

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 $n\geq N$,

$$ \frac{|x_{n}-2|}{2|x_{n}|}<\frac{2\epsilon}{2}=\epsilon. $$

Answer 2

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$( n_0,n_1)$:

$|1/x_n-1/2| < (1/2)|x_n-2| <(1/2)\epsilon < \epsilon.$

Answer 3

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.