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 $$\forall \epsilon>0 (\exists \delta >0(\forall x,y \in [a,\infty] (\left|x-y\right|<\delta \implies \left|f(x)-f(y)\right|< \epsilon)))$$


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 $\lim_{n\to \infty} (f(x_n)-f(y_n))=0$.

  • $\begingroup$ What is the issue that you want discuss here? $\endgroup$
    – Leon
    Jan 4 at 9:56
  • $\begingroup$ In the last expressions, you should use absolute value! $\endgroup$
    – Leon
    Jan 4 at 9:58
  • $\begingroup$ Because in the last inequality, we don't know $x_n\geq y_n$ or $y_n\geq x_n$. $\endgroup$
    – 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 $\lim_{n\to \infty} (f(x_n)-f(y_n))=0$.


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