L'Hôpital with the Fundamental Theorem of Calculus

When to reach for this

Some konkur limits put an integral with a variable upper limit into a quotient, e.g.

$\lim_{x\to a}\frac{\displaystyle\int_{c}^{x} g(t)\,dt}{h(x)}.$

You usually cannot find a closed form for $\int_c^x g(t)\,dt$ (the integrand may have no elementary antiderivative). The escape is to never integrate at all: if the quotient is indeterminate, apply L'Hôpital's rule and differentiate the integral with the Fundamental Theorem of Calculus (FTC), which hands you the integrand back with no work.

The two tools

FTC, Part 1 (Leibniz form). If $g$ is continuous near $a$ and $F(x)=\displaystyle\int_{c}^{x} g(t)\,dt$, then

$F'(x)=g(x).$

With a moving upper limit $u(x)$ you also need the chain rule:

$\frac{d}{dx}\int_{c}^{u(x)} g(t)\,dt = g\bigl(u(x)\bigr)\,u'(x).$

L'Hôpital's rule. If $\dfrac{N(x)}{D(x)}$ is of the form $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ as $x\to a$, and the right-hand limit exists, then

$\lim_{x\to a}\frac{N(x)}{D(x)} = \lim_{x\to a}\frac{N'(x)}{D'(x)}.$

The magic is that $N(x)=\int_c^x g(t)\,dt$ has $N'(x)=g(x)$ — the integral sign is gone after a single differentiation.

The algorithm

1. Check the form first. A variable-limit integral $\int_c^x g\,dt\to 0$ as $x\to c$ whenever $g$ is bounded near $c$ (so the area shrinks to zero). Confirm the denominator $\to 0$ too. Without a genuine $\tfrac00$ or $\tfrac\infty\infty$, L'Hôpital is illegal.
2. Differentiate top and bottom. Numerator: use FTC, $\frac{d}{dx}\int_c^{u(x)} g\,dt = g(u(x))\,u'(x)$. Denominator: ordinary differentiation.
3. Evaluate the new quotient by substituting $x=a$. Knowing the integrand's limit $g(x)\to g(a)$ is enough — you never needed the antiderivative.
4. Repeat or expand if still indeterminate. If $\tfrac00$ persists, apply L'Hôpital again or replace the integrand by its leading Taylor term.

Mini-example

Evaluate $\displaystyle\lim_{x\to 0}\frac{\int_{0}^{x} e^{t^{2}}\,dt}{x}$ (the numerator has no elementary antiderivative).

• Form: as $x\to 0$ the integrand $e^{t^2}$ is bounded near $0$, so $\int_0^x e^{t^2}\,dt\to 0$, and the denominator $\to 0$. It is $\tfrac00$. ✓
• Differentiate: numerator $\to e^{x^{2}}$ by the FTC; denominator $\to 1$.
• Evaluate: $\displaystyle\lim_{x\to 0}\frac{e^{x^{2}}}{1}=e^{0}=1.$

A harder integrand changes nothing: only its value at $x=a$ matters after differentiating.

Pitfall: Always verify the indeterminate form before differentiating. If the integrand $\to L\neq 0$, then $\int_0^x g\,dt \approx Lx$ for small $x$, which already tells you the leading behavior — but you may only invoke L'Hôpital when both numerator and denominator genuinely tend to $0$ (or both to $\infty$). And don't forget the chain-rule factor $u'(x)$ when the upper limit is $u(x)$ rather than a bare $x$.