L'Hôpital's Rule

L'Hôpital's Rule

When a limit has the indeterminate form $0/0$ or $\infty/\infty$, L'Hôpital's rule lets you evaluate it using derivatives:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

provided the right-hand limit exists and $g'(a) \neq 0$.

Classic Examples

$\lim_{x \to 0} \frac{\sin x}{x} = \frac{\cos 0}{1} = 1$

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \frac{2x}{1}\bigg|_{x=1} = 2$

$\lim_{x \to 0} \frac{e^x - 1}{x} = \frac{e^0}{1} = 1$

Numerical Implementation

Numerically, we approximate $f'(a)$ and $g'(a)$ using central differences with a small $h$:

double lhopital(double (*f)(double), double (*g)(double),
                double x0, double h) {
    double df = (f(x0 + h) - f(x0 - h)) / (2.0 * h);
    double dg = (g(x0 + h) - g(x0 - h)) / (2.0 * h);
    return df / dg;
}

This avoids the $f(a)/g(a)$ $0/0$ problem by working with the derivatives directly.

When to Apply It

L'Hôpital applies when you have $0/0$ or $\pm\infty/\infty$. It does not apply to other forms like $0 \cdot \infty$ or $1^\infty$ directly — those need algebraic manipulation first.

Your Task

Implement double lhopital(f, g, x0, h) that returns $f'(x_0) / g'(x_0)$ using central differences.