Implementation of exp function in C using Taylor Series expansion

I am trying to write code to calcultate ex using:

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$

This is the code I have, which works for small values of x.
Please do not pay attention to the variable e, specifically to the fact that user enters it.

#include 
#include 

int main(void) {
    double x0,x1,e,x,sum=0.0;
    int n=0;
    scanf("%lf %lf",&x,&e);
    x1=1;
    do{
        sum+=x1;
        x0=x1;
        x1=(x/++n)*x0;
    }
    while(x1/x0>e);

    printf("sum=%lf, \nexp=%lf",sum,exp(x));
    return 0;
}

For input 50 0.00001 number of decimal places for e doesn't matter.
it generates:

sum=5184705528587076239360.000000, 
exp=5184705528587072045056.000000

For input 21 0.00000001:

sum=1318815734.483214, 
exp=1318815734.483215

The difference only increases as x increases.
I assume that condition is wrong but see no other way of doing it.

Termination condition not correct

To get the precision you request, you should stop the loop when x1 <= e, not when x1 / x0 <= e. If you do that, you can now get rid of x0 because it no longer serves a purpose. I adjusted your program to use the new termination condition and it gave me the results I believe you are looking for. I also renamed e to precision because it was confusing for a variable to be named e when it didn't actually hold the value of the constant \$e\$.

#include 
#include 

int main(void) {
    double x1,precision,x,sum=0.0;
    int n=0;
    scanf("%lf %lf",&x,&precision);

    // Sums: 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...
    //
    // To get an approximation of e^x.
    //
    // Stops when the next term to add becomes smaller than precision.
    x1=1;
    do {
        sum += x1;
        x1  *= (x/++n);
    } while (x1 > precision);

    printf("sum=%lf,\nexp=%lf\n",sum,exp(x));
    return 0;
}

StackExchange Code Review Q#123970, answer score: 2