Indeterminate forms

What "indeterminate" really means. When you evaluate $\lim_{x\to a}f(x)$ by plugging in $x=a$, the result is usually a genuine number and the limit is settled instantly. But sometimes the plug-in produces an expression that carries no information about the actual limiting value — at Class 11 level, the one that matters is

where: the numerator $p(x)\to 0$ and the denominator $q(x)\to 0$ simultaneously as $x\to a$, for a ratio $f(x)=\dfrac{p(x)}{q(x)}$. The phrase "no information" is doing real work here: $\tfrac00$ is not a value the way $\tfrac{3}{2}$ or $0$ is a value. It is a statement that the method of "just substitute" has broken down at this particular point, and a different route to the same limit must be found.

Why $\tfrac{0}{0}$ shows up at all. If a polynomial $p(x)$ satisfies $p(a)=0$, the factor theorem guarantees $(x-a)$ divides $p(x)$ exactly. So if both $p(a)=0$ and $q(a)=0$, then $(x-a)$ is a common factor buried inside both the numerator and the denominator. For every $x\neq a$ — and remember, a limit only ever cares about values near $a$, never the value exactly at $a$ — that common factor can be cancelled, leaving a simplified expression whose limit at $x=a$ is perfectly ordinary and substitutable. The indeterminate form is therefore a disguise, not a dead end: something is hiding inside the fraction that algebra can strip away.

The three standard unmaskings. Depending on how the fraction is built, one of three techniques exposes and removes that hidden common factor:

• Factorisation — for polynomial or rational expressions, factor $p(x)$ and $q(x)$ directly and cancel the shared $(x-a)$.
• Rationalisation — when a surd (square-root) term is producing the zero, multiply numerator and denominator by the conjugate so the radical clears and a cancellable factor appears.
• Trigonometric identity plus standard limit — for trigonometric expressions, rewrite using identities (half-angle formulas, $\tan x=\sin x/\cos x$, and so on) until the expression reduces to a multiple of the two standard results $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{\tan x}{x}=1$.

The figure below shows the shape of the situation graphically: a curve that has a genuine "hole" at $x=a$ rather than a true discontinuity — the function is undefined exactly at the point in question, but the graph approaches one clear height from both sides, and that height is the limit the indeterminate form was hiding.

Crucially, an indeterminate form is a comment on a method failing at a point, not a verdict on the limit itself. Almost every $\tfrac00$ situation met at this level does resolve to a well-defined finite number once the hidden factor is removed.

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\tan x-\sin x}{x^3}$, identifying and resolving the indeterminate form using trigonometric identities.

Step 1. Try direct substitution at $x=0$: numerator $=\tan 0-\sin 0=0-0=0$, denominator $=0^3=0$. This is $\tfrac00$ again, but here neither factorisation nor rationalisation applies directly, since the expression is trigonometric — an identity-based rewrite is needed instead.

Step 2. Rewrite $\tan x=\dfrac{\sin x}{\cos x}$ and factor out $\sin x$:

Then use the half-angle identity $1-\cos x=2\sin^2(x/2)$:

Step 3. Rewrite $\sin x=2\sin(x/2)\cos(x/2)$ so every piece is built from $x/2$:

Dividing by $x^3=8(x/2)^3$ and regrouping toward the standard-limit shape:

Step 4. As $x\to 0$, also $x/2\to 0$, so $\lim_{x\to 0}\dfrac{\sin(x/2)}{x/2}=1$ by the standard limit, while $\cos(x/2)\to 1$ and $\cos x\to 1$. Substituting these:

The indeterminate $\tfrac00$ at the start concealed the finite value $\tfrac12$, reached this time not by cancelling a polynomial factor but by rewriting toward the two standard trigonometric limits.

Competitive stretch

The same three-technique toolkit extends smoothly: a limit like $\lim_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}$ gives $\tfrac00$ on substitution, but rationalising by multiplying through by $\sqrt{1+x}+1$ turns the numerator into $x$, cancels against the denominator, and leaves $\lim_{x\to 0}\dfrac{1}{\sqrt{1+x}+1}=\tfrac12$ — the rationalisation technique doing for surds exactly what factorisation does for polynomials.

• Writing $\tfrac{0}{0}=1$ or $\tfrac{0}{0}=0$ as a final answer — an indeterminate form is not a number and cannot be the answer itself; it only marks that more work is needed.
• Confusing $\tfrac00$ with a fraction like $\tfrac{5}{0}$: if the denominator alone vanishes while the numerator stays nonzero, the function is undefined or unbounded there — a genuinely different situation, not an indeterminate form.
• Cancelling a factor from numerator and denominator without first checking, by direct substitution or the factor theorem, that both truly vanish at the same point.
• Stopping after one failed technique instead of working through the sequence — factorise, then rationalise, then try trigonometric identities — until one actually removes the indeterminacy.

Ask the tutor: got a $\tfrac00$ limit and can't tell which of the three techniques to reach for first — want to walk through it together?