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A limit in math determines the value that a function f(x) approaches as x gets closer and closer to a certain value.

Let me use the equation f(x)=x^2/x as an example.
Obviously, f(x) is undefined if x is 0 (x=0, x2=0, 0/0 is undefined).
But, what happens when we calculate f(x) as x approaches 0?

x=0.1, f(x)=0.01/0.1 = 0.1
x=0.01, f(x)=0.0001/0.01 = 0.01
x=0.00001, f(x)=0.0000000001/0.00001 = 0.00001
We can easily see that f(x) is approaching 0 as x approaches 0.

What about when f(x)=1/x^2, and we approach 0?

x=0.1, f(x)=1/0.01=100
x=0.01, f(x)=1/0.0001=10000
x=0.00001, f(x)=1/0.0000000001=10000000000
As x approaches 0, f(x) approaches positive infinity.

You will be given two things in whatever input format you like:

  • f(x) as an eval-able string in your language
  • a as a floating point

Output the value that f(x) approaches when x approaches a. Do not use any built-in functions that explicitly do this.

Input Limitations:

  • The limit of f(x) as x approaches a will always exist.
  • There will be real-number values just before and just after f(a): you will never get f(x)=sqrt(x), a=-1 or anything of the sort as input.

Output Specifications:

  • If f(x) approaches positive or negative infinity, output +INF or -INF, respectively.
  • If f(a) is a real number, then output f(a).

Test Cases:

f(x)=sin(x)/x, a=0; Output: 1
f(x)=1/x^2, a=0; Output: +INF
f(x)=(x^2-x)/(x-1), a=1; Output: 1
f(x)=2^x, a=3; Output: 8

Shortest code wins. Good luck!

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  • \$\begingroup\$ What would you like to have displayed in the case of f(x)=1/x and a=0 \$\endgroup\$ Commented Oct 15, 2012 at 12:01
  • \$\begingroup\$ @Matt: You will never get that input. The limit of f(x) as x approaches a will always exist. \$\endgroup\$ Commented Oct 15, 2012 at 23:58
  • \$\begingroup\$ A mathematician wouldn't say "the limit exists" in a case like 1/x² | ₓ→₀, but I suppose it's clear what you mean. \$\endgroup\$ Commented Oct 17, 2012 at 22:38
  • \$\begingroup\$ @leftaroundabout: What would they call it then? Indeterminate? :) I'd like to know. \$\endgroup\$ Commented Oct 18, 2012 at 0:23
  • \$\begingroup\$ At least in standard analysis they'd just say "the function diverges for x → 0". The destinction between positive and negative infinity is quite cumbersome (and hardly worth it) when doing proper mathematical proofs. \$\endgroup\$ Commented Oct 18, 2012 at 0:37

2 Answers 2

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C, 402 bytes

Plenty of room for golfing, just proof of concept to begin with :P

Compiles under clang, osx.

Note: For linux you may have to change the code to use something like: "ld -shared f.o -o f.so -lm"

#include<dlfcn.h>
#include<stdio.h>
float e,r;
main(int i,char**v){
    // create and compile temporary c file for 'f(x)'
    // note: include offset within function
    FILE*s=fopen("f.c","w");
    fprintf(s,"#include<math.h>\nfloat f(float x){x+=%s;return %s;}",v[2],v[1]); 
    fclose(s);
    system("cc -fPIC -c f.c;ld -bundle f.o -o f.so -lm");
    float(*f)(float)=dlsym(dlopen("f.so",0),"f");

    for(i=0;1;e=(e==0)?1e-37:e*2){
        r=f(e);
        if(isinf(r))i=1;else if(!isnan(r)){
            i?printf("%cINF\n",r>0?43:45):printf("%f\n",r);
            break;
        }
    }
}

Run as:

./a.out "sin(x)/x" 0
./a.out "1/(x*x)" 0
./a.out "(x*x-x)/(x-1)" 1
./a.out "pow(2,x)" 3

Results:

1.000000
+INF
1.000000
8.000000
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Python 146 185

Figured I'd post the obvious answer before anyone else does. Note, reads two lines from stdin with f(x) on the first and a on the second.

Edit: Made it actually work, as per beary's comment Edit 2: i is much smaller, so there are much fewer inputs that it should fail for.

from math import*
f,a,i=eval("lambda x:"+input()),eval(input()),2e-308
while 1:
 try:
     r=str(f(a+i)).upper()
     if'NAN'!=r:break
 except:
     i*=-2
print('I'==r[0]and'+'+r or r)
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  • \$\begingroup\$ Can you put in the input format? Also, you seem to have neglected the sign on the INF (it should be +inf or -inf). I really don't mind "obvious" answers, as long as it works without any functions calculating it. \$\endgroup\$ Commented Oct 15, 2012 at 4:32
  • 1
    \$\begingroup\$ This is probably being overly picky, but this does not work for f(x) = 1/x, a = -1*10^-161 and f(x) = 1/(x-1*10^-161), a = 0 \$\endgroup\$ Commented Oct 15, 2012 at 12:08
  • \$\begingroup\$ Darn, I guess I'll have to make i the smallest float possible, which should bandadge those problems. \$\endgroup\$ Commented Oct 15, 2012 at 12:21
  • \$\begingroup\$ python provides a machine epsilon as sys.float_info.epsilon, which is probably more accurate for what you're attempting \$\endgroup\$ Commented Oct 15, 2012 at 22:18
  • \$\begingroup\$ Thanks for that ardnew, though I think you meant sys.float_info.min? i in my code is nearly that on my machine (saving a few characters, sacrificing portability). \$\endgroup\$ Commented Oct 16, 2012 at 14:00

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