Real Analysis 5 | Sandwich Theorem [dark version]

Q: What algebraic trick was used to handle c_n = √(n^2 + 1) − n?

Multiplying by the conjugate √(n^2 + 1) + n. This converts the expression into a rational form: (√(n^2 + 1) − n)(√(n^2 + 1) + n) = (n^2 + 1) − n^2 = 1. The result is c_n = 1 / (√(n^2 + 1) + n).

Q: How does the example produce the squeeze bounds 0 and 1/n?

The expression c_n = 1 / (√(n^2 + 1) + n) is nonnegative, so c_n ≥ 0. Also, √(n^2 + 1) + n > n, so 1 / (√(n^2 + 1) + n) < 1/n. Thus 0 ≤ c_n ≤ 1/n, and since 1/n → 0, the sandwich theorem gives c_n → 0.

TL;DR

The sandwich theorem says that if a_n ≤ c_n ≤ b_n eventually and both a_n and b_n converge to the same limit L, then c_n also converges to L.

Briefing Cornell Notes

Briefing

The sandwich theorem for limits turns hard-to-evaluate expressions into something manageable by trapping them between two sequences that share the same limit. If two convergent sequences a_n and b_n satisfy a_n ≤ c_n ≤ b_n for all sufficiently large n, and if both a_n and b_n approach the same number L, then c_n must also converge to L. This matters because many limit problems produce a “middle” term whose limit is unclear directly, but becomes obvious once it’s squeezed between bounds that are already understood.

The proof strategy starts by reducing the problem to a sequence that should converge to 0. With a_n and b_n converging to the same limit a, the difference b_n − a_n shrinks to 0. Define d_n = c_n − a_n, so d_n sits between 0 and b_n − a_n. Since b_n − a_n becomes arbitrarily small in absolute value beyond some index N (the epsilon definition of convergence), d_n is forced to be smaller than any chosen epsilon as well. That pins down d_n → 0. Adding back a_n then transfers the limit from d_n to c_n, giving c_n → a.

A key practical note: the inequalities don’t need to hold for every single n. It’s enough that they hold “eventually,” meaning for all n beyond some point. Since limits ignore finitely many early terms, the squeeze still determines the same limiting behavior.

The transcript then applies the theorem to a concrete example: c_n = √(n^2 + 1) − n. Direct limit rules can’t be applied immediately because both √(n^2 + 1) and n grow without bound, and the subtraction hides the true scale. The fix is an algebraic trick: multiply by the conjugate √(n^2 + 1) + n. This removes the square root and the minus sign at once:

√(n^2 + 1) − n = [(n^2 + 1) − n^2] / [√(n^2 + 1) + n] = 1 / [√(n^2 + 1) + n].

Because √(n^2 + 1) is always positive, the denominator is greater than n, so the fraction is always less than 1/n. The expression is also nonnegative, so c_n is trapped between 0 and 1/n. Both bounds converge to 0, so the sandwich theorem forces c_n to converge to 0.

Overall, the core takeaway is a workflow: (1) identify two sequences that bound the target term, (2) confirm both bounds approach the same limit, and (3) use eventual inequalities to conclude the middle term shares that limit. In problems where subtraction or radicals obscure the limit, rewriting the expression to reveal such bounds is often the decisive step.

Cornell Notes

The sandwich theorem provides a reliable way to find limits when a term is hard to evaluate directly. If a_n ≤ c_n ≤ b_n for all sufficiently large n and both a_n and b_n converge to the same limit L, then c_n must also converge to L. The proof reduces the problem to showing a related sequence d_n = c_n − a_n is squeezed between 0 and b_n − a_n, which goes to 0. Once d_n → 0, adding back a_n gives c_n → a. The theorem is especially useful when algebraic manipulation (like using a conjugate) turns a complicated expression into something that can be bounded by simpler sequences.

What exact conditions make the sandwich theorem work for limits?

Two sequences a_n and b_n must satisfy a_n ≤ c_n ≤ b_n for all n beyond some index (eventually). If lim_{n→∞} a_n = lim_{n→∞} b_n = L, then lim_{n→∞} c_n = L. The “eventually” part means the inequalities can fail for finitely many initial terms without changing the limit.

Why does the proof introduce a new sequence d_n = c_n − a_n?

Because subtracting a_n centers the squeeze around 0. With a_n ≤ c_n ≤ b_n and both a_n and b_n approaching the same limit a, the difference d_n = c_n − a_n satisfies 0 ≤ d_n ≤ b_n − a_n. Since b_n − a_n → 0, the squeeze forces d_n → 0.

How does the epsilon definition enter the sandwich theorem proof?

Convergence of b_n − a_n to 0 means: for any ε > 0, there exists N such that for all n ≥ N, |b_n − a_n| < ε. Because d_n is trapped between 0 and b_n − a_n, this inequality implies d_n < ε for n ≥ N, which matches the definition that d_n → 0.

What algebraic trick was used to handle c_n = √(n^2 + 1) − n?

Multiplying by the conjugate √(n^2 + 1) + n. This converts the expression into a rational form: (√(n^2 + 1) − n)(√(n^2 + 1) + n) = (n^2 + 1) − n^2 = 1. The result is c_n = 1 / (√(n^2 + 1) + n).

How does the example produce the squeeze bounds 0 and 1/n?

The expression c_n = 1 / (√(n^2 + 1) + n) is nonnegative, so c_n ≥ 0. Also, √(n^2 + 1) + n > n, so 1 / (√(n^2 + 1) + n) < 1/n. Thus 0 ≤ c_n ≤ 1/n, and since 1/n → 0, the sandwich theorem gives c_n → 0.

Review Questions

In your own words, what changes if the inequalities a_n ≤ c_n ≤ b_n hold only for n large enough rather than for all n?
Given bounds a_n ≤ c_n ≤ b_n with lim a_n = lim b_n = L, what must lim c_n equal?
For c_n = √(n^2 + 1) − n, what conjugate should be multiplied to simplify the expression, and what final form makes bounding possible?

Key Points

1
The sandwich theorem says that if a_n ≤ c_n ≤ b_n eventually and both a_n and b_n converge to the same limit L, then c_n also converges to L.
2
Only eventual inequality matters; finitely many violations do not affect the limit.
3
A common proof move is to define d_n = c_n − a_n so that 0 ≤ d_n ≤ b_n − a_n.
4
Once b_n − a_n → 0, the squeeze forces d_n → 0, and adding back a_n yields c_n → a.
5
When radicals and subtraction obscure a limit, multiplying by a conjugate can turn the expression into a form that’s easy to bound.
6
For c_n = √(n^2 + 1) − n, rewriting gives c_n = 1 / (√(n^2 + 1) + n), which is squeezed between 0 and 1/n, so the limit is 0.

Highlights

If two sequences squeeze a third term from below and above and both approach the same limit, the middle term must approach that same limit.

The proof reduces the problem to showing a squeezed nonnegative sequence d_n goes to 0, then reconstructs c_n by adding back a_n.

Multiplying by the conjugate √(n^2 + 1) + n turns √(n^2 + 1) − n into 1 / (√(n^2 + 1) + n).

Bounding 0 ≤ c_n ≤ 1/n immediately yields lim c_n = 0.

Topics

Sandwich Theorem
Limits
Convergent Sequences
Epsilon Proofs
Conjugate Trick