Finding a monotonically increasing function with limit 1

Question

To polish/improve a homework answer, I am trying to find a monotonically, continuous, strictly increasing function $f$ with these properties:

1. $f(0) = 0$
2. $\lim_{x \to \infty} f(x) = 1$

(I don't care what happens when $x < 0$.)

This task seems harder than I thought. My first instinct was to use something related with logs, but the problem there is that $\log x$, while it has a similar shape as the function I'm hoping to derive, will exceed 1.

My next thought was to use $f(x) = 0$ if $x = 0$ and $f(x) = \frac{x-1}{x}$ if $x > 0$ (and undefined if otherwise), but the problem is that when $x < 1$, we can get negative values, and if I try to set another case, it's difficult for me to ensure that, for instance, $f(0.09) < f(1.01)$.

Does anyone have any advice? And furthermore, does anyone have strategies on how to create functions satisfying certain properties should I need to do these things in the future?

Answer 1

Another, more natural example (to me anyway) is $$f(x)=1-\frac{1}{x^2+1}.$$ This has the advantage that the limit to $\pm \infty$ is $1$.

Answer 2

How about $f(x) = \tanh{x}$, which satisfies those properties.

$$\tanh{x} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

Answer 3

Hints:

If you know that derivable functions are monotone increasing if their derivative is positive and

$$\left(\frac{x-1}{x}\right)'=\left(1-\frac{1}{x}\right)'=\frac{1}{x^2}$$

then you're done, otherwise you'll have to work harder.

Of course, in your case defining $\,f(0)=0\,$ doesn't bother at all.