Indeterminate forms and asymptotic reasoning

What makes a form "indeterminate." An expression like $\frac00$ is not a number — it is a placeholder for "the limit of a ratio whose numerator and denominator both vanish." The actual limit depends on the rate at which each piece approaches its boundary value, and that rate is exactly what a derivative measures. This is why calculus, not algebra, is the right tool.

The rule itself. Suppose $f,g$ are differentiable near $a$ (except possibly at $a$), $g'(x)\neq0$ near $a$, and the pair $f,g$ both tend to $0$ together, or both tend to $\pm\infty$ together, as $x\to a$. Then, provided the limit on the right exists (or is $\pm\infty$),

where: $f,g$ are the numerator and denominator functions, $a$ is the point (or $\pm\infty$) the limit is taken at, and $f',g'$ are their derivatives computed independently — not via the quotient rule.

Why it is true. Cauchy's mean value theorem states that for $f,g$ differentiable on an interval with $g'\neq0$, there is a point $c$ between $a$ and $x$ with $\dfrac{f(x)-f(a)}{g(x)-g(a)} = \dfrac{f'(c)}{g'(c)}$. In the $\frac00$ case, $f(a)=g(a)=0$, so the left side collapses to exactly $f(x)/g(x)$. As $x\to a$, the trapped point $c$ is squeezed toward $a$ too, so $f'(c)/g'(c) \to \lim_{x\to a} f'(x)/g'(x)$ — which is the rule. If the new ratio $f'/g'$ is itself $\frac00$ or $\frac\infty\infty$, the same argument applies again: L'Hôpital's rule may be chained.

Funneling the other five forms in. The remaining indeterminate shapes are all algebraic disguises of $\frac00$ or $\frac\infty\infty$:

For $0\cdot\infty$: rewrite $fg$ as $\dfrac{f}{1/g}$ (a $\frac00$ form) or $\dfrac{g}{1/f}$ (a $\frac\infty\infty$ form). For $\infty-\infty$: combine into one fraction, e.g. $\dfrac1f-\dfrac1g=\dfrac{g-f}{fg}$, which becomes $\frac00$. For the three exponential forms $0^0,1^\infty,\infty^0$: take logarithms. If $y=f(x)^{g(x)}$, then $\ln y = g(x)\ln f(x)$, a $0\cdot\infty$ form; evaluate $\lim\ln y$ first, then exponentiate to recover $\lim y$. This same machinery is the everyday tool of asymptotic reasoning — pinning down how $f(x)$ behaves as $x\to\infty$ or as $x\to a$, which is exactly what finding asymptotes (a nearby topic in this unit) requires.

Evaluate $\displaystyle\lim_{x\to 0} \frac{x-\sin x}{x^3}$, then use the same log trick to confirm $\displaystyle\lim_{x\to 0^+} x^x = 1$.

Step 1. Check the form at $x=0$: numerator $0-\sin0=0$, denominator $0^3=0$. This is $\frac00$, so differentiate top and bottom separately: $\dfrac{1-\cos x}{3x^2}$.

Step 2. Re-check: at $x=0$ this is again $\dfrac{1-1}{0}=\frac00$, so differentiate once more: $\dfrac{\sin x}{6x}$, still $\frac00$ at $x=0$ — differentiate a third time: $\dfrac{\cos x}{6}$.

Step 3. This last ratio is no longer indeterminate: at $x=0$ it equals $\dfrac{1}{6}$, a genuine finite value. So

Three applications were needed because each intermediate ratio was still $\frac00$ — the form must be re-checked after every differentiation, never assumed.

Step 4. Now the exponential case. Let $y=x^x$ for $x\to0^+$. Take logs: $\ln y = x\ln x$, a $0\cdot(-\infty)$ form since $x\to0$ but $\ln x\to-\infty$. Rewrite as a ratio: $x\ln x = \dfrac{\ln x}{1/x}$, which is $\dfrac{-\infty}{\infty}$ — apply L'Hôpital's rule: $\lim_{x\to0^+}\dfrac{\ln x}{1/x} = \lim_{x\to0^+}\dfrac{1/x}{-1/x^2} = \lim_{x\to0^+}(-x) = 0$.

Step 5. So $\lim_{x\to0^+}\ln y = 0$, and exponentiating back, $\lim_{x\to0^+} y = e^0 = 1$. Answer: $\lim_{x\to0}\dfrac{x-\sin x}{x^3} = \dfrac16$, and $\displaystyle\lim_{x\to 0^+} x^x = 1$.

Competitive stretch

The same "log first, then exponentiate" pattern proves $\displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x = e$ (a $1^\infty$ form): setting $y=(1+1/x)^x$, $\ln y = x\ln(1+1/x)$ is $\infty\cdot0$; writing it as $\dfrac{\ln(1+1/x)}{1/x}$ (now $\frac00$) and applying L'Hôpital's rule gives $\lim\ln y = 1$, so $\lim y = e$ — worth recognizing on sight rather than re-deriving.

• Applying L'Hôpital's rule to a ratio that is not actually indeterminate — e.g. $\lim_{x\to0}\frac{\sin x}{x+1}$ is $\frac01=0$ directly; differentiating here gives a wrong answer.
• Differentiating numerator and denominator as a single quotient (quotient rule) instead of separately — the rule requires $f'(x)$ and $g'(x)$ computed independently, then divided.
• Stopping after one application when the new ratio is still indeterminate — always re-check the form ($\frac00$ or $\frac\infty\infty$) before declaring the limit.
• Forgetting to exponentiate back ($e^{\lim\ln y}$) after using logarithms for $0^0$, $1^\infty$, or $\infty^0$ — the log-limit is not the final answer.

Ask the tutor: if you're unsure whether your limit is genuinely indeterminate, or which algebraic move turns your specific $0\cdot\infty$ or $\infty-\infty$ case into $\frac00$, ask the AI tutor to check the form with you and walk through the manipulation step by step.