#include
void function(int);
int main()
{
     int x;
     printf(\"Enter x:\");
     scanf(\"%d\", &x);
function(x);
return 0;
}
void functi         
         By default in UNIX, floating-point division by zero does not stop the program with an exception. Instead, it produces a result which is infinity or NaN. You can check that neither of these happened using isfinite.
x = y / z; // assuming y or z is floating-point
if ( ! isfinite( x ) ) cerr << "invalid result from division" << endl;
Alternately, you can check that the divisor isn't zero:
if ( z == 0 || ! isfinite( z ) ) cerr << "invalid divisor to division" << endl;
x = y / z;
With C99 you can use fetestexcept(2) et alia.
This should do it. You need to check for division by zero before performing the division.
void function(int x)
{
    float fx;
    if(x == 0) {
        printf("division by zero is not allowed");
    } else {
        fx = 10/x;
        printf("f(x) is: %.5f",fx);
    }
}
#include<stdio.h>
void function(int);
int main()
{
     int x;
     printf("Enter x:");
     scanf("%d", &x);
function(x);
return 0;
}
void function(int x)
{
    float fx;
    if(x==0) // Simple!
        printf("division by zero is not allowed");
    else
        fx=10/x;            
        printf("f(x) is: %.5f",fx);
}