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The quotient rule needs a nonzero limit in the denominator: $1 / (k + 1)^{- 1} = k + 1$

Statement refuted

Refuted claim: if $x_{k} \to x$ and $y_{k} \to y$ and $x_{k} \neq 0$ for every $k$ , then $y_{k} / x_{k}$ converges.

The witnesses are $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = 1$ : both converge — to $0$ and to $1$ respectively — and every $x_{k}$ is nonzero, but $y_{k} / x_{k} = k + 1$ diverges to $+ \infty$ . The quotient clause of Algebra of limits: sums, scalar multiples, products and quotients requires the limit of the denominator to be nonzero, and this is the extremal failure of exactly that hypothesis. It also instantiates the reciprocal bridge of For positive terms, null and divergence to $+ \infty$ are reciprocal.

Facts & Assumptions

Given: The sequences $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = 1$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F1]

$(x_{k})$ converges to $0$ and every $x_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Reciprocal bridge: for a sequence $(z_{k})$ with $z_{k} > 0$ for every $k$ , $(z_{k})$ converges to $0$ if and only if $(1 / z_{k})$ diverges to $+ \infty$ (For positive terms, null and divergence to $+ \infty$ are reciprocal).

[F3]

Divergence to $+ \infty$ : $z_{k} \to + \infty$ means that for every real $M$ there is $K$ with $z_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F4]

Field arithmetic: every nonzero $u$ has a multiplicative inverse with $u \cdot u^{- 1} = 1$ and $(u^{- 1})^{- 1} = u$ , multiplication is commutative, and $u \cdot 1 = u$ (Field).

[F5]

A sequence diverging to $+ \infty$ is unbounded, and an unbounded sequence is not convergent, the latter being the contrapositive of "convergent implies bounded" (Every convergent sequence is bounded, Divergence to $+ \infty$ and to $- \infty$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F6]

Quotient clause of the algebra of limits: if $(p_{k})$ converges to $p$ , $(q_{k})$ converges to $q$ , and $q \neq 0$ with $q_{k} \neq 0$ for every $k$ , then $p_{k} / q_{k}$ converges to $p / q$ (Algebra of limits: sums, scalar multiples, products and quotients).

[F7]

Every constant real sequence converges to its value: $∣ c - c ∣ = 0 < ε$ at every index for every rational $ε > 0$ (Limits and Cauchy sequences of reals).

Counterexample

technique · direct

1.1givenF1F7

The sequence $(x_{k})$ is positive termwise and converges to $0$ by [F1]; the constant sequence $(y_{k}) = (1)$ converges to $1$ by [F7].

2.1step 1.1F2F3

By the forward direction of the reciprocal bridge [F2], applied to the positive null sequence $(x_{k})$ of step 1.1, the sequence $(1 / x_{k})$ diverges to $+ \infty$ in the sense of [F3].

2.2step 1.1F4

At every index the quotient equals the reciprocal: $y_{k} / x_{k} = 1 \cdot x_{k}^{- 1} = x_{k}^{- 1} = ((k + 1)^{- 1})^{- 1} = k + 1 = 1 / x_{k}$ , using $u \cdot 1 = u$ and the involution $(u^{- 1})^{- 1} = u$ of [F4] at the nonzero element $u = k + 1$ .

3.1step 1.1step 2.1step 2.2F5F6∎

Therefore $(y_{k} / x_{k}) = (1 / x_{k})$ diverges to $+ \infty$ by step 2.1, so by [F5] the quotient is unbounded and not convergent: the refuted claim fails. The quotient clause [F6] is not contradicted — its denominator-limit hypothesis $q \neq 0$ fails here, since the limit of the denominator is $x = 0$ by step 1.1, while its other hypotheses ( $x_{k} \neq 0$ termwise) do hold.

Depends on

Used by

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Sources

John K. Hunter, An Introduction to Real Analysis (complete author-hosted lecture notes, 307 pp.) (standard reference, not scraped)
Jiri Lebl, Basic Analysis I: Introduction to Real Analysis, Volume I (author-hosted complete book PDF) (standard reference, not scraped)

The quotient rule needs a nonzero limit in the denominator: 1/(k+1)−1=k+11/(k+1)−1=k+1