Example 1 — \(\displaystyle \lim_{x\to0}\frac{\sin x}{x}\)
\(0/0\). Apply L'Hôpital: \(\lim_{x\to0}\frac{\cos x}{1}=\cos 0=1.\)
L'Hopital’s Rule Calculator
Compute limx→a f(x)/g(x). If direct substitution gives
0/0 or ∞/∞, apply L’Hôpital:
lim f/g = lim f′/g′ (repeat as needed).
Use math like
sin, cos, exp, log, sqrt, x^2.g(x) must have nonzero derivative at some step.
Use a finite value (e.g., 0, 1, 2).
How many times to apply L’Hôpital (0 = none).
Display precision only.
Helping Notes
- L’Hôpital applies to
0/0or∞/∞forms when f and g are differentiable nearaandg′≠ 0 at the step used. - If direct substitution is defined and finite, that is the limit.
- For two-sided limits with oscillation, the numeric estimate may be unstable; try reducing step size or adjusting expressions.
Result
Equation Preview
If
f(a)/g(a) is not indeterminate, we take that value. Otherwise, we try
f′(a)/g′(a), then higher orders up to k.
Limit Result
If still indeterminate after k steps, we show a cautious numeric estimate using small symmetric steps.
Details & Checks
Derivative strings are symbolic; numeric values are evaluated at
x = a.
What is a L'Hôpital’s Rule Calculator?
A L'Hôpital’s Rule Calculator is a calculus tool that computes limits of ratios producing indeterminate forms like \(0/0\) or \(\infty/\infty\). When direct substitution fails, L'Hôpital’s method replaces the original limit with the limit of derivatives—provided standard hypotheses hold—often transforming a stubborn expression into a simple evaluation. This calculator symbolically differentiates numerator and denominator, checks whether the indeterminate form persists, and, if needed, applies the rule repeatedly. It also assists with related indeterminate forms (\(0\cdot\infty\), \(\infty-\infty\), \(1^\infty\), \(0^0\), \(\infty^0\)) by first rewriting them into a quotient or by using logarithms. All steps are displayed in LaTeX.
About the L'Hôpital’s Rule Calculator
Suppose \(f\) and \(g\) are differentiable on an open interval around \(a\) (possibly excluding \(a\) itself), with \(g'(x)\ne 0\) near \(a\). If \[ \lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0 \quad\text{or}\quad \lim_{x\to a} |f(x)|=\lim_{x\to a} |g(x)|=\infty, \] and if \(\lim_{x\to a}\frac{f'(x)}{g'(x)}\) exists (finite or infinite), then \[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}. \] The same holds for one-sided limits and \(x\to\pm\infty\). The calculator verifies differentiability assumptions where possible, warns about \(g'(x)=0\) near \(a\), and offers series-based hints if repeated applications stall.
How to Use this L'Hôpital’s Rule Calculator
- Enter the functions \(f(x)\) and \(g(x)\) (or a single expression \(\frac{f(x)}{g(x)}\)) and the limit point \(a\) (number, \(\infty\), or \(-\infty\)).
- Compute the direct substitution to detect the form (e.g., \(0/0\), \(\infty/\infty\)).
- If indeterminate, apply L'Hôpital: replace by \(\lim \frac{f'(x)}{g'(x)}\). Repeat until the limit becomes determinate or the method ceases to help.
- For \(0\cdot\infty\), \(\infty-\infty\), or power-type forms, first rewrite into a quotient or take logarithms, then proceed.
- Review the final limit with step-by-step LaTeX and optional checks (series expansion or numeric sampling).
Core Formulas
L'Hôpital’s Rule (basic): \[ \text{If } \frac{f}{g} \text{ is } \frac{0}{0}\text{ or } \frac{\infty}{\infty},\ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \text{ (when the latter exists).} \]
Repeated application: \[ \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{(n)}(x)}{g^{(n)}(x)}\quad \text{if earlier derivatives keep yielding an indeterminate form.} \]
Transform other forms: \[ 0\cdot\infty \Rightarrow \frac{f(x)}{1/g(x)},\quad \infty-\infty \Rightarrow \frac{(a\pm b)}{1}\ \text{combine to a quotient},\quad y^{z}\Rightarrow \ln L=\lim z(x)\ln y(x),\ L=e^{\lim z\ln y}. \]
Examples (Illustrative)
Example 2 — \(\displaystyle \lim_{x\to\infty}\frac{\ln x}{x}\)
\(\infty/\infty\). L'Hôpital: \(\lim_{x\to\infty}\frac{1/x}{1}=0.\)
Example 3 — \(\displaystyle \lim_{x\to0}\frac{e^{x}-1-x}{x^{2}}\)
\(0/0\). First application: \(\lim\frac{e^{x}-1}{2x}\) still \(0/0\). Second: \(\lim\frac{e^{x}}{2}=\tfrac{1}{2}.\)
FAQs
When can I apply L'Hôpital’s Rule?
Only for \(0/0\) or \(\infty/\infty\) forms (after verifying differentiability and \(g'(x)\ne0\) near the point).
Does it work for \(0\cdot\infty\), \(\infty-\infty\), \(1^\infty\), \(0^0\), \(\infty^0\)?
Yes—after rewriting into a quotient or using logarithms to convert power-type forms.
Is L'Hôpital’s Rule valid for one-sided limits or \(x\to\pm\infty\)?
Yes. The theorem holds for one-sided and infinite limits under the same hypotheses.
What if derivatives keep giving an indeterminate form?
Apply repeatedly or switch tactics (algebraic simplification, series/Taylor expansion, or comparison tests).
Can I use it if \(g'(a)=0\)?
Yes, provided \(g'(x)\ne0\) in a punctured neighborhood and the derivative limit exists; otherwise seek another method.
Why did my answer differ from a series solution?
Usually an algebraic slip or applying the rule outside its conditions; recheck steps or compare with Taylor expansion.
How does the calculator handle \(1^\infty\) limits?
It sets \(y(x)^{z(x)}\) to \(e^{z(x)\ln y(x)}\), evaluates \(\lim z\ln y\) via L'Hôpital if needed, then exponentiates.
Is L'Hôpital always faster than algebraic simplification?
No. Factorization or rationalization can be quicker; the tool suggests simpler paths when they exist.
Does using higher derivatives change the limit?
No—if conditions hold and the derivative limit exists, the computed limit equals the original one.
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