L'hopital's Rule Applicator

Math Calculus • Applications of Derivatives

Written by STEM Calculators Team Published December 23, 2025 Updated February 24, 2026

7. L'hopital's Rule Applicator

Evaluate limits of the form \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}\) when the direct substitution gives \(0/0\) or \(\infty/\infty\), by differentiating numerator and denominator (possibly multiple times).

Inputs

Numerator \(f(x)\)

Denominator \(g(x)\)

Supported: + − * / ^, parentheses, variable x, constants pi, e, sin cos tan, ln log(base 10), sqrt abs exp. Implicit multiplication: 2x, (x+1)(x-1), 3sin(x).

Approach

L’Hôpital applies to \(0/0\) or \(\infty/\infty\) forms.

Value \(a\)

You can use pi or e.

Max L’Hôpital applications

Stops if not resolved by this many steps.

Output format

Options Show grid Auto-fit near approach

Notes

Condition checker is numeric at the approach point (or large \(x\) for \(\pm\infty\)). If your limit requires series, algebra, or a different theorem, this tool may say “inconclusive”.

Ready

Graph

Drag to pan • wheel/pinch to zoom • shows \(h(x)=\frac{f(x)}{g(x)}\) near the approach point.

Graph center \(c\)

Auto-fit recommended.

Graph window half-width \(w\)

Graph uses \(x\in[c-w,c+w]\).

x: 0, y: 0, zoom(px/unit): 80 Tip: if values blow up, zoom out or use Auto fit.

Result

Enter \(f(x)\), \(g(x)\), choose the approach, then click Compute.

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7. L'hopital's Rule Applicator — Theory

L’Hôpital’s Rule is a method for evaluating certain limits of quotients \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}\) when direct substitution produces an indeterminate form: \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

\[ \textbf{L’Hôpital’s Rule:}\quad \text{If }\lim_{x\to a}\frac{f(x)}{g(x)} \text{ is } \frac{0}{0}\text{ or }\frac{\infty}{\infty} \text{ and } f,g \text{ are differentiable near }a,\text{ then} \] \[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \quad \text{(provided the right-hand limit exists).} \]

When you are allowed to apply it

The limit must be a quotient \(\frac{f(x)}{g(x)}\).
After substitution \(x=a\) (or as \(x\to\pm\infty\)), the form must be \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
\(f\) and \(g\) must be differentiable near the approach point (except possibly at \(a\)).
\(g'(x)\neq 0\) near the point so the derivative quotient makes sense.

Repeated application

If the first differentiation still gives \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you may apply L’Hôpital again:

\[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f''(x)}{g''(x)} = \cdots \] \[ \text{Stop once the quotient is no longer } \frac{0}{0}\text{ or }\frac{\infty}{\infty}. \]

Worked example

Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin x - x}{x^3}\).

\[ \frac{\sin x - x}{x^3}\xrightarrow[x\to 0]{}\frac{0}{0} \Rightarrow \text{apply L’Hôpital} \] \[ \lim_{x\to 0}\frac{\cos x - 1}{3x^2} \xrightarrow[x\to 0]{}\frac{0}{0} \Rightarrow \text{apply again} \] \[ \lim_{x\to 0}\frac{-\sin x}{6x} \xrightarrow[x\to 0]{}\frac{0}{0} \Rightarrow \text{apply again} \] \[ \lim_{x\to 0}\frac{-\cos x}{6} = -\frac{1}{6}. \]

Common pitfalls

Wrong form: L’Hôpital does not directly apply to \(0\cdot\infty\), \(\infty-\infty\), \(0^0\), \(1^\infty\), \(\infty^0\) unless you first rewrite into a quotient giving \(0/0\) or \(\infty/\infty\).
Differentiability: If \(f\) or \(g\) is not differentiable near the approach point, the rule may fail.
Endless loop: Differentiating repeatedly may never resolve the form; then another method is needed (algebra, series, squeezing, etc.).

Geometric intuition

Roughly, if both \(f\) and \(g\) are heading to 0 (or both heading to infinity), comparing their derivatives compares their local “rates of change,” which can reveal the limit of the ratio.

Frequently Asked Questions

When can I use L'Hôpital's Rule for a limit?

L'Hôpital's Rule applies to a quotient f(x)/g(x) when the limit produces the indeterminate form 0/0 or infinity/infinity as x approaches a value or as x approaches plus/minus infinity. The functions should be differentiable near the approach point so f'(x) and g'(x) are defined.

What does this calculator do when the form is not 0/0 or infinity/infinity?

If the limit is not one of the two required indeterminate forms, L'Hôpital's Rule is not used. The calculator will treat the situation as not applicable and may report that the result is inconclusive for L'Hôpital's Rule.

How many times can L'Hôpital's Rule be applied?

It can be applied repeatedly as long as each new quotient still evaluates to 0/0 or infinity/infinity at the approach. This calculator stops once the form is resolved or when it reaches your Max L'Hôpital applications setting.

Why does the calculator sometimes say "inconclusive"?

The condition check at the approach point is numeric, and some limits require algebraic rewriting, series expansions, or a different theorem to evaluate. If the form does not resolve within the allowed steps or the numeric checks are unstable, the tool may return an inconclusive result.

What is the L'Hôpital formula used by the solver?

If lim f(x)/g(x) is 0/0 or infinity/infinity, the rule replaces it with lim f'(x)/g'(x), provided the new limit exists. The calculator repeats this replacement as needed up to the chosen maximum.