Is there a non-exponential function whose limit at infinity is a real, irrational number?

Question

$e$, for example, can be calculated through a non-polynomial function $(1+1/x)^x$, but I cant think of an example for a non-exponential function (or rational function) where the limit to infinity approaches an irrational constant.

Answer 1

Here's one you know: $\arctan x,$ whose limit at $\infty$ is $\pi/2.$

Answer 2

If $f(x) = 1/x^2 + \pi$, then $\lim_{x \to \infty} f(x) = \pi$ (and same with $\lim_{x \to -\infty} f(x) = \pi$), which is irrational.

Answer 3

It is quite easy to construct such a function from a function, $f(x)$, whose limit at infinity is a finite rational number, $q$. If $r$ is an irrational number, then the limit of $g(x) = f(x) + (r - q)$ is clearly $r$.

Similarly, $h(x) = \frac {f(x)r} q$ also has a limit of $r$ as $x$ goes to infinity.