$\left\{\frac{1}{f(x_n)}\right\}$ converges to $\frac{1}{f(x)}$. Let $f:\Bbb R \to \Bbb R$ be a continuous function. Let $\{x_n\}_{n=1}^\infty$ be a convergent sequence in $\Bbb R$ with $\lim \limits_{n\to\infty}x_n=x$ and $f(x)\ne 0$
I want to show that $\left\{\frac{1}{f(x_n)}\right\}$ converges to $\frac{1}{f(x)}$.
Now It would seem I have:
$$|f(x_n) - f(x)| \lt \epsilon $$
$$|f(x_n) - f(x)| \leq |f(x_n)| - |f(x)| $$
and now I don't get it, I would expect to get some relationship also less than epsilon, times by $-1$ to inverse the epsilon inequality and then find a way to get the ricipricals and that would reverse the epsilon inequality again giving me the result, but I can't see it.
 A: I guess it should be $f(x)\neq 0$. Since $f$ is continuous and $f(x)\neq 0$, $\frac{1}{f}$ is continuous around $x$. Now use that convergent sequences are mapped to convergent sequences under continuous functions.
A: Another idea: with a reasoning like the used by Clément Guérin prove that $x\mapsto\dfrac1x$ is continuous in any point $\ne 0$. Then, use that the composition of continuous functions is continuous.
A: You should not begin with :
$$|f(x_n)-f(x)|$$
try directly :
$$|\frac{1}{f(x_n)}-\frac{1}{f(x)}| $$
It is possible to do so for $n$ "very big" (that is there exists $N$ such that for $n\geq N$, $f(x_n)\neq 0$ because $f(x)\neq 0$).
$$|\frac{1}{f(x_n)}-\frac{1}{f(x)}|=\frac{|f(x_n)-f(x)|}{|f(x_n)f(x)|} $$
Now the numerator will be as small as you want and you can show (I leave it to you) that $|f(x_n)|\geq \frac{|f(x)|}{2}$ for $n$ big enough.
. Then you deduce that :
$$|\frac{1}{f(x_n)}-\frac{1}{f(x)}|\leq \frac{2}{|f(x)|^2}|f(x_n)-f(x)|$$
Then the difference on the left is smaller than a constant multiplied by something as small as you want. 
A: Since $f$ is continuous and $\lim\limits_{n\to \infty}x_n=x$ we have $\lim\limits_{n\to \infty}f(x_n)=f(x)$. That is the sequence $\{f(x_n)\}$ converges. So $\{|f(x_n)|\}$ is. Then $\{|f(x_n)|\}$ is bounded and $m=\min\{|f(x_n)|:n\in \mathbb{N}\}$ exists (finitely).
Let $\epsilon>0$ be arbitrary. Since $\{f(x_n)\}$ converges there exists $n_\epsilon\in \mathbb{N}$ such that for each $n>n_\epsilon$, $|f(x_n)-f(x)|<\epsilon .m.|f(x)|$.
Observe that for each $n>n_\epsilon$,
$\left|\dfrac{1}{f(x_n)}- \dfrac{1}{f(x)}\right|=\dfrac{|f(x_n)-f(x)|}{|f(x_n)||f(x)|}\le \dfrac{|f(x_n)-f(x)|}{m|f(x)|}<\dfrac{\epsilon}{m|f(x)|}.m.|f(x)|=\epsilon $.
Therefore $\left\{\dfrac{1}{f(x_n)}\right\}$ converges to $\dfrac{1}{f(x)}$. $\square$
