# Show that the function $f:[a,\infty] \to \mathbb R$ defined by $x \mapsto \frac{1}{x}$ for $a$ a positive number is uniformly continuous.

Show that the function $$f:[a,\infty] \to \mathbb R$$ defined by $$x \mapsto \frac{1}{x}$$ for $$a$$ a positive number is uniformly continuous.

The function is continuous if and only if $$\forall \epsilon>0 (\exists \delta >0(\forall x,y \in [a,\infty] (\left|x-y\right|<\delta \implies \left|f(x)-f(y)\right|< \epsilon)))$$

$$\left|f(x)-f(y)\right|=\left|\frac{1}{x}-\frac{1}{y}\right|=\frac{\left|x-y\right|}{xy}\le\frac{\left|x-y\right|}{a^{2}}$$

Taking $$\delta= a^2\epsilon$$ follows uniform continuity of $$f$$ over the given interval.

I also used the sequential criterion :

The function is continuous if and only if $$\forall (x_n)_n,(y_n)_n \subset [a,\infty]:\lim_{n\to \infty} (x_n-y_n)=0 \implies \lim_{n\to \infty} (f(x_n)-f(y_n))=0$$

Let $$(x_n)_n,(y_n)_n \subset [a,\infty]$$ be arbitrary sequences with the property that $$\lim_{n\to \infty} (x_n-y_n)=0$$ then $$\lim_{n\to \infty} (f(x_n)-f(y_n)) =\lim_{n\to \infty} \frac{y_n-x_n}{x_ny_n}\le -\frac{1}{a^2}\lim_{n\to \infty} (x_n -y_n)=0$$

But I'm not able to use squeeze theorem to conclude that $$\lim_{n\to \infty} (f(x_n)-f(y_n))=0$$.

• What is the issue that you want discuss here?
– Leon
Jan 4 at 9:56
• In the last expressions, you should use absolute value!
– Leon
Jan 4 at 9:58
• Because in the last inequality, we don't know $x_n\geq y_n$ or $y_n\geq x_n$.
– Leon
Jan 4 at 10:03

Note that $$\lim a_n=0\Leftrightarrow \lim |a_n|=0$$.
The last expression should be $$\lim_{n\to \infty} |f(x_n)-f(y_n)|=\lim_{n\to \infty} \frac{|y_n-x_n|}{x_ny_n}\le \frac{1}{a^2}\lim_{n\to \infty} |x_n -y_n|=0.$$
Now, the squeeze theorem allows us to conclude that $$\lim_{n\to \infty} (f(x_n)-f(y_n))=0$$.