AP Calculus AB and BC

When the Limit Laws Cannot Be Applied

The limit laws let you split a limit into pieces only when each piece has a limit of its own. Run them backwards and they break: the limit of a sum can exist while neither part has a limit at all, as sin of one over x plus its negative shows.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   Each piece has a limit at a, on its own

   $$\lim_{x \to a} f(x) \text{ and } \lim_{x \to a} g(x) \text{ both exist}$$

   The law is an equation between two sides, and the right side has to make sense before the equation means anything. Reading the law from right to left is fine; reading it from left to right, by splitting a limit you already know exists, assumes the answer.

   Drop it and the theorem fails

   $$\sin\!\left(\tfrac{1}{x}\right) + \left(-\sin\!\left(\tfrac{1}{x}\right)\right) = 0$$

   sin(1/x) and its negative, added together, near x = 0

   The sum is identically zero, so its limit at the origin is 0 without any difficulty. But the check confirms that $\sin(1/x)$ has no limit at 0: it keeps swinging across the full range from $-1$ to 1 however close you look. Splitting this limit into two pieces produces two expressions that do not exist, and no amount of algebra recovers the 0 you started with.

2. 2

   For the quotient law, the denominator's limit is nonzero

   $$\lim_{x \to a} g(x) \ne 0$$

   Division by the limit is only defined when that limit is not zero. When it is zero the law says nothing, which is exactly the situation that produces the indeterminate forms $0/0$ and calls for factoring, conjugates, or L'Hopital's Rule. The law is not violated in that case, it simply does not apply, so no counterexample is claimed.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. Let $L$ and $M$ be the two limits, and fix a tolerance $\varepsilon > 0$.
2. Choose a radius small enough that $|f(x) - L| < \varepsilon/2$, and another small enough that $|g(x) - M| < \varepsilon/2$. Both radii exist because both limits were assumed to exist.
3. Inside the smaller radius, the triangle inequality gives $|(f+g) - (L+M)| \le |f - L| + |g - M| < \varepsilon$.
4. Since $\varepsilon$ was arbitrary, the sum tends to $L + M$. Every step used the existence of both limits separately, which is why the law cannot be run in reverse.

What it does not say