Show that if $f$ is continuous on $(0,1)$ and $\lim_{x \to 0}f(x) = + \infty$ then $f$ is not uniformly continuous

Question

I must show that if $f$ is continuous on $(0,1)$ and $$\lim_{x \to 0}f(x) = + \infty$$ then $f$ is not uniformly continuous. I am not sure if we have to use epsilon- delta proofs or how to show this.

Answer 1

Assume that $|x-x'|<\delta \implies |f(x) - f(x')|<1$. Then if $|x-x'|<2$ then $|f(x) - f(x')|<\frac 2\delta$.

Consider $|\epsilon|<1$.

Then $$|f(\epsilon)| \le |f(1)| + |f(1) - f(\epsilon)|\le |f(1)|+ \frac 2\delta $$ So $f$ is bounded around 0;

Answer 2

a uniformly continous function should convert cauchy sequences into cauchy sequences; this function convert the cauchy sequence (1/n) into an unbounded sequence then not a cauchy one.

Answer 3

Uniformly continuous means that whenever I tell you a positive number $\epsilon$, you can (in principle) find a positive number $\delta$ so that whenever you move from one point to another on the $x$-axis, as long as you move less than $\delta$ away from where you started, the function value changes by less than $\epsilon$.

Now, assume the function $f$ is uniformly continuous but with $\lim_{x\to0} f(0) = \infty$. Let's say I give you $\epsilon =1$, and you find a $\delta$ that works. Now, start at $x=\frac12$, and take $\delta$-sized steps to the left. For each step, the function value cannot change by more than $1$. But when you've taken enough steps to be within reach of $0$, the next step can take you to an $x$ with as large function value as you want, clearly contradicting the $\epsilon=1$ bound.

Therefore, a uniformly continuous function cannot behave like that.