Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e: Visual walkthrough

We are proving one line: $lim_{x \to 0} \frac{s i n x}{x} = 1, x measured in == radians == .$

Everything below earns each piece of that line.

Step 1 — Draw the stage: a unit circle and one angle

WHAT. Draw a circle of radius $1$ centred at the origin $O$ . Mark the point $B = (1, 0)$ where the circle crosses the positive $x$ -axis. Sweep an angle $x$ counter-clockwise and land on the circle at point $P$ .

WHY. A circle of radius exactly $1$ is chosen so that lengths and angles become the same number — that is the whole trick, and Step 2 explains it. Everything we compare will live inside this one picture, so no symbol will appear without a place to point at.

PICTURE. Look at the blue circle. $O$ is the centre, $B$ the red dot on the axis, $P$ the orange dot on the rim. The little arc from $B$ to $P$ (green) is what the angle "cuts out".

Figure — Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Step 2 — What "radian" means, and why $arc = x$

WHAT. Define the measure of $x$ to be the length of the green arc $B P$ . That definition is what "radians" means.

WHY. We need $x$ (an angle, a rotation) and a length (something we can compare to $sin x$ ) to be the same quantity. Radian measure makes that literally true: on a circle of radius $r$ , arc length $= r θ$ , and with $r = 1$ we get arc $= x$ . If we used degrees, arc $= \frac{π}{180} x$ instead — a different number — and the clean " $= 1$ " would break. This is the single reason the whole result needs radians.

PICTURE. The green arc is labelled with its own length: it is $x$ . Same curve, two readings — "angle" and "length."

Step 3 — Read off $sin x$ and $tan x$ as heights

WHAT. Drop a vertical line from $P$ straight down to the $x$ -axis; its height is $sin x$ . Now go the other way: draw the vertical tangent line to the circle at $B$ , and extend ray $O P$ until it hits that tangent line at a point $T$ . The height of $T$ above the axis is $tan x$ .

WHY. These two heights are the two "competitors" we will squeeze $x$ between. We need geometric objects, not formulas, so let us define both directly on the picture:

$sin x = vertical drop from the rim height of P, tan x = ray extended to the tangent height of T .$

PICTURE. Orange segment $= sin x$ (short, ends on the circle). Red segment $= tan x$ (taller, ends on the outside tangent line). Notice by eye: the orange height is smaller than the arc, and the arc is smaller than the red height. That ordering is the entire proof — Steps 4–6 make it exact.

Step 4 — Three nested regions (the sandwich bread)

WHAT. Look at three regions that sit inside one another:

inner triangle $O B P$ (vertices $O$ , $B$ , $P$ ),
circular sector $O B P$ (the pie slice bounded by the two radii and the arc),
outer triangle $O B T$ (vertices $O$ , $B$ , $T$ ).

Because $P$ is on the circle and $T$ is outside it, region 1 $\subset$ region 2 $\subset$ region 3.

WHY. Areas are easy to compute and they respect containment: if one shape sits inside another, its area is no bigger. So containment hands us an inequality for free, with no trig identities needed. That inequality will trap $x$ .

PICTURE. Three shaded layers on the same figure: filled orange (inner triangle), the green pie slice on top of it, the pale red outer triangle behind. The nesting is visible — each larger region contains the previous.

Step 5 — Turn the three regions into three areas

WHAT. Compute each area.

inner △ \frac{1}{2} sin x \leq sector \frac{1}{2} x \leq outer △ \frac{1}{2} tan x .

Term by term:

WHY. We swap the pictures (nested shapes) for numbers (areas), keeping the same $\leq$ ordering from Step 4. Now we can do algebra.

PICTURE. Same three regions, each labelled with its area. The bars on the right compare the three magnitudes at a small angle so you can see $\frac{s i n x}{2} < \frac{x}{2} < \frac{t a n x}{2}$ .

Step 6 — Algebra: fold the sandwich around $\frac{s i n x}{x}$

WHAT. Start from Step 5 and clean up.

Multiply all three parts by $2$ : $sin x \leq x \leq tan x .$

Divide through by $sin x$ . For $0 < x < \frac{π}{2}$ we have $sin x > 0$ , so dividing by a positive number keeps every $\leq$ pointing the same way: $1 \leq \frac{x}{s i n x} \leq \frac{t a n x}{s i n x} = \frac{1}{c o s x} .$ (Here $\frac{tan x}{sin x} = \frac{sin x / cos x}{sin x} = \frac{1}{cos x}$ — the $sin x$ cancels.)

Take reciprocals. Flipping $a \leq b \leq c$ of positive numbers into $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$ reverses each $\leq$ , so: $cos x \leq \frac{s i n x}{x} \leq 1.$

WHY. We wanted $\frac{s i n x}{x}$ alone in the middle, trapped between two things we understand. Each move (multiply, divide by positive $sin x$ , reciprocate) is a legal inequality step, and we said out loud why the direction stayed or flipped.

PICTURE. A number line: the value $\frac{s i n x}{x}$ (orange tick) is caught between the floor $cos x$ (blue) and the ceiling $1$ (gray). As $x$ shrinks the floor rises toward $1$ and the gap slams shut.

Step 7 — Let $x \to 0$ : the Squeeze Theorem closes the trap

WHAT. Send $x \to 0$ . The ceiling is the constant $1$ . The floor is $cos 0 = 1$ , so $cos x \to 1$ . Both walls converge to $1$ : $\to 1 cos x \leq \frac{s i n x}{x} \leq = 1 1 .$

By the Squeeze (Sandwich) Theorem, anything trapped between two things that both go to $1$ must itself go to $1$ : $x \to 0^{+} lim \frac{sin x}{x} = 1.$

WHY. A limit sandwiched between two known limits inherits their common value — that is exactly what the Squeeze Theorem is for, and it is precisely the tool this shape hands us.

PICTURE. The floor $cos x$ and ceiling $1$ plotted as curves; the orange curve $\frac{s i n x}{x}$ threads between them and all three meet at the single point $(0, 1)$ .

Step 8 — The other side and the exact centre (no case left out)

WHAT. We only proved the right limit ( $x \to 0^{+}$ ). Check the two remaining scenarios.

Negative side, $x \to 0^{-}$ . The function $\frac{s i n x}{x}$ is even: replace $x$ by $- x$ , $\frac{s i n ( - x )}{- x} = \frac{- s i n x}{- x} = \frac{s i n x}{x} .$ So the graph is a mirror image across the vertical axis — the left limit equals the right limit. Hence the two-sided limit is $1$ .

Exactly $x = 0$ . The expression is $\frac{s i n 0}{0} = \frac{0}{0}$ : undefined, a hole in the graph. The limit is $1$ , but the value at $0$ does not exist. Limits describe the approach, not the landing point.

WHY. The parent claim is a two-sided limit; a one-sided proof is only half the job. Covering $x < 0$ and the degenerate $x = 0$ point is what makes the boxed result airtight.

PICTURE. Full graph of $\frac{s i n x}{x}$ across negative and positive $x$ : symmetric hump peaking at $1$ , with an open circle (hole) drawn exactly at $(0, 1)$ .

Recall Quick check on the cases

Is $\frac{s i n x}{x}$ defined at $x = 0$ ? ::: No — it is $0/0$ , a hole. Only the limit is $1$ . Why does $x \to 0^{-}$ give the same answer? ::: The function is even, so its graph is mirror-symmetric about the $y$ -axis; left and right limits match.

The one-picture summary

Everything compressed: the nested inner triangle $\subset$ sector $\subset$ outer triangle give $sin x \leq x \leq tan x \Rightarrow cos x \leq \frac{s i n x}{x} \leq 1 x \to 0 1.$

Recall Feynman retelling — the whole walkthrough in plain words

Draw a circle of radius one and open a small angle. Because we measure angles by the arc they cut on this circle, the angle number is exactly the arc length — that is why we insist on radians. Now stand three shapes next to each other: a skinny triangle inside the pie slice, the pie slice itself, and a fatter triangle outside it. Since each shape is packed inside the next, their areas line up as $\frac{s i n x}{2} \leq \frac{x}{2} \leq \frac{t a n x}{2}$ . Clean up the algebra — multiply by two, divide by the positive $sin x$ , flip reciprocals — and $\frac{s i n x}{x}$ gets pinned between $cos x$ and $1$ . Shrink the angle: the floor $cos x$ rises to $1$ , the ceiling is already $1$ , so the poor thing in the middle has nowhere to go but $1$ . The graph is a mirror image on the left, and right at zero there is just a tiny hole — but the value it heads toward from both sides is $1$ . Done.

Connections

Squeeze (Sandwich) Theorem — the closing move of Step 7.
Radian Measure — the reason arc $= x$ (Step 2) and sector $= \frac{1}{2} x$ (Step 5).
Derivative of sin and cos — this limit is $\frac{d}{d x} sin x$ at $0$ .
Indeterminate Forms — why $\frac{s i n 0}{0} = \frac{0}{0}$ needs a real argument.
L'Hôpital's Rule — would need $sin^{'}$ , which needs this limit, so it is circular here.
Taylor Series — $sin x = x - \frac{x ^{3}}{6} + \dots$ makes $\frac{s i n x}{x} = 1 - \frac{x ^{2}}{6} + \dots \to 1$ .