If a sequence of real numbers converges to 1, then what do we say about its reciprocal?

Suppose the sequence of real number $\lbrace x_n \rbrace _{n=1} ^{\infty}$ converges to 1, $x_n \neq 0 \; \; \; \forall n=1,2,3,...$. Show that $\lbrace \frac{1}{x_n} \rbrace _{n=1} ^{\infty}$ converges to 1 using episilon-delta definition.

Hint: If the sequence of real number $\lbrace x_n \rbrace _{n=1} ^{\infty}$ converges to $B \in \mathbb{R} | \lbrace 0 \rbrace$, and if $x_n \neq 0 \; \; \; \forall n=1,2,3,...$ then
1. $\lbrace \frac{1}{x_n} \rbrace _{n=1} ^{\infty}$ converges to $\frac{1}{B}$

1. $\vert x_n \vert > \frac{\vert B \vert }{2}$ for some $n>N$

These were my thoughts to prove this idea: If the sequence of real number $\lbrace x_n \rbrace _{n=1} ^{\infty}$ converges to $1 \in \mathbb{R} | \lbrace 0 \rbrace$, and if $x_n \neq 0 \; \; \; \forall n=1,2,3,...$ then given an $\epsilon > 0$ there exist an $N \in \mathbb{ Z} ^{+}$ such that $\vert x_n -1 \vert < \epsilon$ whenever $n>N$.

But we know that $\vert x_n -1 \vert= \vert 1 - x_n \vert =\geq \vert 1 \vert - \vert x_n \vert= 1-\vert x_n \vert$

It therefore follows that $1-\vert x_n \vert < \epsilon$ whenever $n>N$.

Now choose $\epsilon = \frac{1}{2}$, then we have $1-\vert x_n \vert < \frac{1}{2}$ whenever $n>N$ from this we have $\vert x_n \vert > \frac{1}{2}$ whenever $n>N$

From here one can proceed as pewani has given below: that is why I accepted his answer below...

• What did you try? – 5xum Aug 25 '15 at 11:58
• Might help to note that convergent sequences are bounded. – Santeri Aug 25 '15 at 12:05
• I tried similar way as Pewani's answer below. And it works. Good luck. – Patrick Chidzalo Aug 25 '15 at 12:20
• Guys tell me why you are down voting this que? – Patrick Chidzalo Aug 25 '15 at 16:03
• Please reopen this question. I have verified that and the answer given by pewani below is 100 percent correct. Read K.G. Binmore for more information. – Patrick Chidzalo Aug 25 '15 at 16:11

Suppose $\lbrace x_n \rbrace$ converges to $1$, then given $\epsilon > 0$, there exists an $N > 0$ such that $\vert x_n -1 \vert< \epsilon$ whenever $n > N$.
Now we assume that $x_n \neq 0$, then $\vert x_n \vert > \frac{1}{2}$
and hence $\displaystyle \vert \frac{1}{x_n} -1 \vert=\vert \frac{1-x_n}{x_n} \vert <2 \vert 1-x_n \vert < 2 \epsilon$