Calculus I · Patterns That Go Somewhere

Limits that fight back

Everything so far has been gentle: substitute, simplify, read off the answer. The limits that teach you something are the ones where that fails. Look at these three:

$$n\left(\sqrt{1+\tfrac1n}-1\right), \qquad \sqrt[n]{n}=n^{1/n}, \qquad \left(1+\frac1n\right)^{n}.$$

Try the naive move on the first. As $n\to\infty$, the factor $n$ runs off to infinity while the parenthesis $\sqrt{1+\frac1n}-1$ collapses to $0$. So the answer is… infinity times zero? That is not a number. It is a fight between a part pulling the product up and a part pulling it down, and “$\infty\cdot 0$” is just our shorthand for “the winner is not yet decided — look closer.”

These deadlocks have a name: indeterminate forms. The usual suspects are

$$\infty\cdot 0,\qquad \frac{\infty}{\infty},\qquad \frac{0}{0},\qquad 1^{\infty},\qquad \infty-\infty,\qquad 0^{0}.$$

When substitution lands you on one of these, you have learned exactly one thing: you cannot stop here. The limit might be $0$, might be $\infty$, might be any finite number in between — it depends on how fast each part moves, and finding out is the whole skill. Our three sequences are, in order, an $\infty\cdot0$, an $\infty^0$, and a $1^\infty$. Let’s win all three. First we need one new tool.

Toolkit: The Squeeze Theorem

If a sequence is trapped between two others heading to the same place, it has no choice but to go there too.

Squeeze Theorem. If $a_n \le b_n \le c_n$ for all large $n$, and both $a_n \to L$ and $c_n \to L$, then $b_n \to L$.

Use it when $b_n$ itself is hard but you can bound it above and below by things you understand. The art is choosing good walls.

A quick win, to watch it move: what is $\displaystyle\lim_{n\to\infty}\dfrac{\sin n}{n}$? The numerator $\sin n$ lurches around forever and has no limit of its own — but it never escapes the corridor $-1 \le \sin n \le 1$. Divide through by $n$:

$$-\frac1n \;\le\; \frac{\sin n}{n} \;\le\; \frac1n.$$

Both walls, $\pm\frac1n$, go to $0$. Squeezed between them, $\dfrac{\sin n}{n}\to 0$ — and we never had to tame the wild numerator. We just fenced it in.

◆ Proof: the Squeeze Theorem

Play the game. Given $\varepsilon>0$: since $a_n\to L$ there’s a cutoff $N_1$ past which $a_n > L-\varepsilon$; since $c_n\to L$ there’s a cutoff $N_2$ past which $c_n < L+\varepsilon$. Let $N$ be the larger of the two. Then for every $n>N$, the sandwich $a_n \le b_n \le c_n$ gives $$L-\varepsilon < a_n \le b_n \le c_n < L+\varepsilon,$$ so $|b_n - L| < \varepsilon$. Beaten for every $\varepsilon$, so $b_n\to L$. $\blacksquare$

Fight one: $n\left(\sqrt{1+\tfrac1n}-1\right)$ (an $\infty\cdot0$)

The move with an indeterminate form is almost always to rewrite it until a shape you can finish. Here the troublemaker is the difference of a square root and $1$, and the standard cure is to rationalize — multiply by the conjugate $\sqrt{1+\tfrac1n}+1$ over itself, which turns the difference of roots into a clean difference of the things under them:

$$n\left(\sqrt{1+\tfrac1n}-1\right) = n\cdot\frac{\left(1+\tfrac1n\right)-1}{\sqrt{1+\tfrac1n}+1} = n\cdot\frac{1/n}{\sqrt{1+\tfrac1n}+1} = \frac{1}{\sqrt{1+\tfrac1n}+1}.$$

The $n$ canceled the $\tfrac1n$ — the two combatants annihilated, leaving something tame. As $n\to\infty$, $\tfrac1n\to0$, so the bottom heads to $\sqrt{1}+1 = 2$, and

$$n\left(\sqrt{1+\tfrac1n}-1\right) \to \frac12.$$

Not $0$, not $\infty$, but exactly $\tfrac12$: the growing $n$ and the shrinking parenthesis were perfectly matched, and they balance at one-half. (Tuck this away: in Chapter 3 you’ll recognize $\tfrac12$ as the slope of $\sqrt{x}$ at $x=1$. You are computing a derivative right now without knowing it — and we’ll meet the same $\frac00$-after-rationalizing move again the moment we reach functions in Chapter 2.)

Fight two: $\sqrt[n]{n}$ (an $\infty^0$, and a genuine surprise)

This one should bother you. Take the $n$-th root of $n$: $\sqrt[2]{2}\approx1.41$, $\sqrt[3]{3}\approx1.44$, $\sqrt[10]{10}\approx1.26$, $\sqrt[100]{100}\approx1.047$, $\sqrt[1000]{1000}\approx1.0069$. The thing under the root is racing to infinity, yet the root is settling toward $1$. The form is $\infty^0$ — a huge base, a vanishing exponent $\frac1n$ — and the vanishing exponent wins. (Intuition: roots are ferocious levelers; a giant number rooted hard enough is tamed almost to $1$.) The proof is a small gem.

◆ Proof: $n^{1/n}\to 1$

Since $n^{1/n}\ge1$, write $n^{1/n}=1+h_n$ with $h_n\ge0$; we must show $h_n\to0$. Raise to the $n$-th power: $n=(1+h_n)^n$. Expand by the binomial theorem and throw away every term but one — keep only the $k=2$ term, legal because all terms are nonnegative: $$n = (1+h_n)^n \;\ge\; \binom{n}{2}h_n^2 = \frac{n(n-1)}{2}\,h_n^2.$$ Divide by $\frac{n(n-1)}{2}$: $\;h_n^2 \le \dfrac{2}{n-1}$, so $0 \le h_n \le \sqrt{\dfrac{2}{n-1}}$. The right wall is built from $\frac{1}{n-1}\to0$, so it $\to0$; the left wall is $0$. Squeeze: $h_n\to0$, hence $n^{1/n}=1+h_n\to1$. $\blacksquare$ The trick worth stealing: when a binomial expansion is all-positive, keeping a single well-chosen term is a valid lower bound — and one term is often all you need.

Making the theorem earn its living: a sequence that defines itself

The monotone-convergence principle feels abstract until you meet a sequence with no formula at all, built by feeding each term back into itself:

$$a_1=\sqrt2,\qquad a_{n+1}=\sqrt{2+a_n}.$$

So $a_2=\sqrt{2+\sqrt2}\approx1.848$, $a_3\approx1.962$, $a_4\approx1.990$, climbing toward… something. We can’t write $a_n$ in closed form — but we can still find the limit exactly, and this is the principle’s finest hour.

◆ Layer 2 — Solving a self-defined sequence

Step 1 — bounded above by $2$ (induction): $a_1=\sqrt2<2$; and if $a_n<2$, then $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=2$. So every term stays under $2$. Step 2 — increasing: $a_{n+1}>a_n$ means $\sqrt{2+a_n}>a_n$, i.e. (square, both sides positive) $2+a_n>a_n^2$, i.e. $a_n^2-a_n-2<0$, i.e. $(a_n-2)(a_n+1)<0$ — true for every $a_n$ with $0<a_n<2$, which Step 1 guarantees. Step 3 — therefore it converges (increasing and bounded above) to some limit $L$. Step 4 — pin down $L$. Take the limit of both sides of the rule $a_{n+1}=\sqrt{2+a_n}$. The left side $\to L$; the right side $\to\sqrt{2+L}$. So $L$ must obey its own equation: $$L=\sqrt{2+L}\;\Longrightarrow\;L^2=2+L\;\Longrightarrow\;(L-2)(L+1)=0.$$ Hence $L=2$ (reject $-1$: the terms are positive). The sequence we couldn’t even write down converges to exactly $2$. That move — let the limit satisfy the rule — is your first fixed point: a value a process leaves unchanged. It returns whenever you study how an iteration settles, including, in spirit, the iterations of Chapter 8.

All 12 chapters of Calculus I · Foundations · All courses