# A function that escapes to infinity with a finite input

I was wondering whether there exists a function that escapes to infinity with a finite input. For a specific example, how about $f(0)=0$ and as $x$ tends to $10$, $f(x)$ tends to infinity. The use of this would be to produce unimaginably large numbers with imaginable inputs.

If there are many such functions, I would prefer a formula which is short and snappy. If need be, magical operators such as an infinite sum or product would suffice. I understand that the tangent function has many poles (is that what you call an un-definition?), but they are sort of expensive to compute. I am not adept in calculus, but if you must... and at the very least I would like the function to be computable.

I am just a bit curious, is all. Can anyone help me? Perhaps the answer is obvious.

• What about things such as $\frac{1}{10-x}-\frac{1}{10}$ ? – Claude Leibovici Feb 1 '14 at 9:37
• I thought a reciprocal would be in order. Nice and simple, an answer. So, I assume that replacing the constant $10$ with another number $n$ would generate a function that has a pole at $n$. So simple. I feel so dumb. – bimmo Feb 1 '14 at 9:42
• In fact, you can select a function $f(x)$ which goes to infinity for $x=n$ and which has a finite value for $x=0$. Then use $g(x)=f(x)-f(0)$ – Claude Leibovici Feb 1 '14 at 9:51
• consider $y=1/(n-x)-1/n$. I have realised that decreasing the numerator of the current reciprocals reduces the bend in the curve. (I have Graphmatica up). A numerator of 0 produces a flat line at y=0 and a hole in that line at y=n. The gradient at y=0 seems to dictate this bend. – bimmo Feb 1 '14 at 10:13
• You can also use $y=(n-x)^{-a}-n^{-a}$ using $a>0$. If I amy ask, what do you plan to do ? If I can help, my pleasure. – Claude Leibovici Feb 1 '14 at 10:19

Look at $\tan(x)$. https://www.wolframalpha.com/input/?i=tan%28x%29. As $x \to \frac\pi2$ then $f(x) \to \infty$. Also true for any integer multiple of $\frac\pi2$.
In general, functions of the form $f(x) = \frac{g(x)}{h(x)}$ will tend to infinity as $h(x)$ tends to 0, which doesn't necessarily require $x$ to tend to $\infty$. Disclaimer: $g(x)$ and $h(x)$ have to be well defined on the same interval, continuous and $h$ doesn't divide $g$ for all $x$ for this to be the case.
Here's an example that fits your criteria $f(0) = 0$ and $f(x) \to \infty$ as $x \to 10$: $f(x) = \tan(\frac{\pi x}{20})$ If you want to construct a function $g$ where $g(x) \to \infty$ as $x \to k$ for some non-zero real number $k$, you can simply use $g(x)=\frac1{x-k}$.
• I have a graph up. It seem that using either @Claude's reciprocal function or this tangent function gives a non-zero gradient at $f(0)$. Interesting. This tangent function is 'smoother' or 'softer' than the reciprocals, presumably because of it's greater slope at $f(0)$. Is the non-zero slope necessary? Would pushing it to 0 produce a square 'kick' to infinity at f(k)? – bimmo Feb 1 '14 at 10:05