Special Limits: sin(x)/x and (1 + 1/n)^n

Learning Objectives

By the end of this section, you will be able to:

1. State and intuitively justify the two fundamental limits $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\displaystyle\lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^{n}=e$.
2. Derive the squeeze inequality $\cos x \le \sin x / x \le 1$ from the unit circle and apply the Squeeze Theorem to prove the first limit.
3. Use the sinc limit to evaluate many related limits — including $(1-\cos x)/x^{2}$, $\tan x / x$, and $\sin(kx)/x$ — without invoking L'Hôpital.
4. Connect the compound-interest sequence $(1+1/n)^{n}$ to the continuous limit $\lim_{h\to 0}(1+h)^{1/h}$ and to $(e^{h}-1)/h \to 1$.
5. Verify both limits numerically in plain Python and recognize them as the limits that define $(\sin x)' = \cos x$ and $(e^{x})' = e^{x}$ in PyTorch's autograd.

Why These Two Limits Are Special

Every standard limit rule we have met so far — the sum, product, quotient, and composition laws — says if you already know the pieces, here is how to combine them. But somewhere the chain has to bottom out. Somewhere a few non-trivial limits must be computed from first principles, and all the other calculus rules will then ride on top of them.

Two limits play exactly this role throughout all of analysis. Everything you know about the derivative of sine, cosine, tangent, e^x, ln, and every decay, oscillation, or compound-growth process in physics, economics, and machine learning can be traced back to these two sentences:

sin(x)/x — Geometric Intuition

Draw a unit circle centered at O and pick a small positive angle $x$ (in radians). Three natural lengths appear in the picture:

• The vertical drop from the circle point to the horizontal axis has length $\sin x$.
• The arc on the circle from the horizontal axis to that point has length exactly $x$ — that is the geometric definition of a radian.
• The tangent segment dropped from the end of the arc to the horizontal axis (outside the circle) has length $\tan x$.

Comparing areas of the three shaded regions in the interactive below:

Divide the whole chain by $\tfrac{1}{2}\sin x > 0$ to get

and invert (all three quantities are positive for $0<x<\pi/2$):

As $x\to 0$, $\cos x\to 1$ (cosine is continuous at 0), so the ratio is trapped between two things that both converge to 1. The Squeeze Theorem forces $\sin x / x$ to converge to 1 as well.

Interactive: Squeeze sin(x)/x onto 1

Drag the slider (or press Play) to shrink the angle $x$. Watch the green, violet, and red regions flatten onto each other in the unit circle and — at the same time — the purple curve $\sin x / x$ get pinched onto the horizontal line $y=1$.

The Sandwich Proof

Worked Example: Why sin(5x)/x → 5

Compute $\displaystyle\lim_{x\to 0}\dfrac{\sin(5x)}{x}$ without L'Hôpital, Taylor series, or any machinery beyond the first special limit.

Playground: Limits Built from sin(x)/x

Pick a formula and shrink the window. Each curve has a removable hole at 0; the green line marks the value the hole would fill in. Notice how every curve flattens onto its green line as $x\to 0$, and how the numerical probe table loses two decimal digits per row — a signature of algebraic rate-of-convergence.

(1 + 1/n)^n — The Compound-Interest Story

In 1683 Jacob Bernoulli asked a practical question that secretly founded modern analysis: if a bank pays 100% interest once a year, you end up with $2 on each $1. What if they compound twice a year — 50% every six months?

Four times a year (25% per quarter) gives $(1+\tfrac{1}{4})^{4}=\$2.44\ldots$. Monthly gives $(1+\tfrac{1}{12})^{12}=\$2.613\ldots$. The more often you compound, the more you earn — but the payoff of adding another compounding step keeps shrinking. Is there a ceiling? Yes: the sequence $a_n=(1+1/n)^{n}$ climbs forever but never crosses

This number e is the one Euler later placed at the center of analysis. Its definition is literally the limit of the compound-interest sequence.

e = compounding gains in the limit of instantaneous compounding. Every appearance of $e^{t}$ in physics — radioactive decay, RC circuit voltage, population growth, Planck radiation — is the continuous limit of a discrete multiplicative process compounded ever more finely.

Interactive: Watch the Balance Climb to e

The left panel shows the running balance of $1 compounded $n$ times a year. The right panel plots $a_n=(1+1/n)^{n}$ on a logarithmic axis. Drag the slider to change $n$ by orders of magnitude.

Why Does It Converge?

Unlike the sinc limit, there is no simple geometric picture — the proof is about numbers on a line, not shapes in a plane. The standard argument is that the sequence is monotonically increasing and bounded above. A classical theorem then forces it to converge.

Step 1 — Monotonicity via the binomial theorem

Expand by the binomial formula:

Each factor $(1-j/n)$ strictly increases with $n$ and the series has one extra positive term when $n$ grows by 1. Every summand grows, so the whole sum grows: $a_{n+1}>a_{n}$.

Step 2 — Boundedness via factorial comparison

Each product $\prod(1-j/n)\le 1$, so

(We used $k!\ge 2^{k-1}$ for $k\ge 1$.) So $a_n<3$ for all $n$. A bounded increasing sequence must converge. Call the limit $e$. The series on the right also converges (to the same number) — that is the other standard formula $e=\sum_{k\ge 0}1/k!$.

From Discrete n to Continuous x

The same number $e$ appears in three equivalentlimits — the discrete sequence we just studied and two continuous-variable versions that come up constantly in calculus:

The bridge from the discrete to the continuous form is simply the substitution $n=1/h$: as $n\to\infty$, $h\to 0^{+}$, and

The continuous version $(1+h)^{1/h}\to e$ works for $h\to 0$ from either side, not just through integer $n$.

Worked Example: Continuous Compounding on $1000

A bank offers 5% APR on a $1000 deposit for one year. They will compound at any frequency you choose. How much money do you have at the end?

Python: Watching Both Limits Converge

Before jumping to PyTorch, let us see the two limits converge in the simplest possible Python. No libraries beyond math. Click any line on the right panel — the left panel shows exactly what value every variable and argument carries at that moment.

1import math
2
3def sinc_ratio(x):
4    """f(x) = sin(x) / x — defined for x != 0, limit at 0 is 1."""
5    return math.sin(x) / x
6
7def euler_sequence(n):
8    """a_n = (1 + 1/n)^n — the compound-interest sequence converging to e."""
9    return (1 + 1 / n) ** n
10
11# --- sin(x)/x as x -> 0 ---
12print("sin(x) / x ->  1")
13print(f"{'x':>10}  {'f(x)':>18}  {'|f(x) - 1|':>14}")
14print("-" * 46)
15for x in [1.0, 0.1, 0.01, 0.001, 0.0001]:
16    f = sinc_ratio(x)
17    print(f"{x:>10.4e}  {f:>18.12f}  {abs(f - 1):>14.2e}")
18
19# --- (1 + 1/n)^n as n -> infinity ---
20print()
21print("(1 + 1/n)^n ->  e")
22print(f"e = {math.e:.15f}")
23print(f"{'n':>12}  {'a_n':>18}  {'e - a_n':>14}")
24print("-" * 48)
25for n in [1, 10, 100, 10_000, 1_000_000]:
26    a = euler_sequence(n)
27    print(f"{n:>12,d}  {a:>18.12f}  {math.e - a:>14.2e}")

sin(x) / x ->  1
         x                f(x)       |f(x) - 1|
----------------------------------------------
1.0000e+00    0.841470984808    1.59e-01
1.0000e-01    0.998334166468    1.67e-03
1.0000e-02    0.999983333417    1.67e-05
1.0000e-03    0.999999833333    1.67e-07
1.0000e-04    0.999999998333    1.67e-09

(1 + 1/n)^n ->  e
e = 2.718281828459045
           n                 a_n         e - a_n
------------------------------------------------
           1      2.000000000000    7.18e-01
          10      2.593742460100    1.25e-01
         100      2.704813829422    1.35e-02
      10,000      2.718145926825    1.36e-04
   1,000,000      2.718280469156    1.36e-06

PyTorch: Using These Limits as Derivatives

The classical proof that $(\sin x)' = \cos x$ uses $\sin h/h\to 1$. The proof that $(e^{x})'=e^{x}$ uses $(e^{h}-1)/h\to 1$. PyTorch's autograd implements those derivative rules, so we can verify numerically that the two "special" limits are exactly the numbers autograd produces.

1import torch
2
3# ---- sin'(0) = 1 follows from sin(x)/x -> 1 ----
4x = torch.tensor(0.0, requires_grad=True)
5y = torch.sin(x)
6y.backward()
7print(f"d/dx sin(x) at x=0 : {x.grad.item():.8f}")  # expect 1.0
8
9# ---- (eˣ)'(0) = 1 follows from (eᴴ - 1)/h -> 1 ----
10x2 = torch.tensor(0.0, requires_grad=True)
11y2 = torch.exp(x2)
12y2.backward()
13print(f"d/dx exp(x) at x=0 : {x2.grad.item():.8f}")  # expect 1.0
14
15# ---- Verify the discrete limit via a finite difference ----
16h = torch.tensor(1e-4)
17fd_sin = torch.sin(h) / h                         # -> 1
18fd_exp = (torch.exp(h) - 1) / h                   # -> 1
19print(f"sin(h)/h        at h=1e-4 : {fd_sin.item():.8f}")
20print(f"(exp(h)-1)/h    at h=1e-4 : {fd_exp.item():.8f}")

Where Both Limits Show Up

Common Pitfalls

Summary

Two limits — one trigonometric, one exponential — act as the base cases for almost every derivative rule in single-variable calculus. Everything else is built from them by algebraic manipulation.

Key Takeaways

1. sin x / x → 1 is the geometric fact that the sine of a tiny angle is the angle (measured in radians). It is proved with one squeeze and ends up under every derivative of every trig function.
2. (1 + 1/n)^n → e is the analytic fact that compounding converges. Equivalent to $(e^{h}-1)/h\to 1$ and $\ln(1+h)/h\to 1$, so it underlies every derivative involving e and ln.
3. Both limits are of indeterminate form. Neither follows from arithmetic alone — one needs geometry, the other needs a monotone-bounded argument.
4. Numerically, sinc converges quadratically, the e-sequence only linearly. "Same limit" in analysis does not mean "same speed" in computation.
5. Every gradient PyTorch computes for $\sin, \cos, \exp, \ln$ is these two limits in disguise. Autograd doesn't take a new limit each time — it uses the closed-form rule that these limits proved once and for all.

Coming Next: With these two special limits in our toolbox, we can finally formalize the shortcut every student wants for $0/0$ and $\infty/\infty$ forms — L'Hôpital's rule. The next section shows how to use it, when it fails, and why it's a corollary (not a replacement) of the work we did here.