Order Limits and the Squeeze Theorem

Limits respect order, with one subtle exception. These facts, together with the limit laws, compute most limits without returning to $\varepsilon$ every time.

Idea clave

Order: if $a_n \le b_n$ for all $n$ (or all large $n$), $a_n \to a$ and $b_n \to b$, then $a \le b$. Squeeze: if $a_n \le c_n \le b_n$ for all large $n$ and $a_n \to L$ and $b_n \to L$, then $c_n \to L$. Strict inequalities can become equalities in the limit.

Por qué funciona

Order. Suppose instead $a > b$, and let $\varepsilon = \frac{a - b}{2}$. Eventually $a_n > a - \varepsilon$ and $b_n < b + \varepsilon$, and $a - \varepsilon = b + \varepsilon$, so $a_n > b_n$, a contradiction. Squeeze. Let $\varepsilon > 0$. Eventually both $L - \varepsilon < a_n$ and $b_n < L + \varepsilon$, and then $L - \varepsilon < a_n \le c_n \le b_n < L + \varepsilon$, so $|c_n - L| < \varepsilon$. $\blacksquare$

The squeeze theorem is how analysts handle anything bounded times something that goes to $0$: since $-1 \le \sin n \le 1$, $\left|\frac{\sin n}{n}\right| \le \frac{1}{n} \to 0$. It also handles sums of many small terms, by bounding every term by the smallest and the largest.

Worked examples

Ejemplo resuelto: A bounded wiggle

Find $\lim \frac{3n + 2\sin n}{4n + 1}$.

Divide top and bottom by $n$: $\frac{3 + 2\frac{\sin n}{n}}{4 + \frac{1}{n}}$.
$-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}$, so by the squeeze theorem $\frac{\sin n}{n} \to 0$.

Respuesta: $\frac{3}{4}$ by the limit laws.

Ejemplo resuelto: A sum with $n$ terms

Find $\lim \sum_{k=1}^{n} \frac{n}{n^{2} + k}$.

Each of the $n$ terms is between $\frac{n}{n^{2} + n}$ (the smallest, at $k = n$) and $\frac{n}{n^{2} + 1}$ (the largest, at $k = 1$).
So $\frac{n^{2}}{n^{2} + n} \le \sum_{k=1}^{n} \frac{n}{n^{2} + k} \le \frac{n^{2}}{n^{2} + 1}$, and both bounds tend to $1$.

Respuesta: $1$

Ejemplo resuelto: A root of a sum

Find $\lim \left(2^{n} + 5^{n}\right)^{1/n}$.

$5^{n} \le 2^{n} + 5^{n} \le 2 \cdot 5^{n}$, so $5 \le (2^{n} + 5^{n})^{1/n} \le 2^{1/n} \cdot 5$.
$2^{1/n} \to 1$, so both bounds tend to $5$.

Respuesta: $5$: the larger base wins.

Ejemplo resuelto: Where a limit can land

Suppose $a_n \to L$ and $2 \le a_n \le 5$ for every $n \ge 10$. What can you say about $L$?

Apply the order theorem to $2 \le a_n$ and to $a_n \le 5$ (the first nine terms do not matter).
So $2 \le L \le 5$. The ends are possible: $5 - \frac{1}{n} \to 5$ stays in $[2, 5]$.

Respuesta: $L \in [2, 5]$, a closed interval.

Error común

Keeping a strict inequality in the limit. From $a_n < b_n$ you may conclude only $\lim a_n \le \lim b_n$. Example: $0 < \frac{1}{n}$ for every $n$, but both sides tend to $0$. In particular, a sequence of positive numbers can have limit $0$. Check any "$<$" in a limit argument against the example $\frac{1}{n}$.

Worked examples

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