Why a Geometric Comparison Gives a Limit
The quotient \(\frac{\sin x}{x}\) is undefined at \(x=0\), so direct substitution cannot evaluate its limit there. The key is to compare the quotient with expressions that do have a clear limit. A unit circle provides exactly the needed comparison: for a small positive angle, a triangle fits inside the circular sector, and the sector fits inside a larger triangle formed using a tangent.
The resulting area inequalities hold for every angle in an interval, not just for selected examples. They will therefore let us use the Squeeze Theorem to establish the limit.
Setting Up the Unit-Circle Comparison
Consider a unit circle with center \(O\), and let \(Q\) be the point \((1,0)\). For an angle \(x\) with \(0<x<\frac{\pi}{2}\), let \(P\) be the point on the circle reached by rotating counterclockwise from \(OQ\) through angle \(x\). Its coordinates are \(P=(\cos x,\sin x)\). Draw the radius \(OP\), the chord \(QP\), and the tangent to the circle at \(Q\). Let \(T\) be where the ray \(OP\) meets that tangent.
The tangent at \(Q\) is the vertical line with equation \(X=1\). Since the ray \(OP\) makes angle \(x\) with the horizontal, its slope is \(\tan x\), so it meets \(X=1\) at \(T=(1,\tan x)\). The triangle \(OQP\) lies inside the sector from \(OQ\) to \(OP\), and that sector lies inside triangle \(OQT\).
The inner triangle has base \(OQ=1\) and height \(\sin x\), so its area is \(\frac{\sin x}{2}\). The sector has radius \(1\) and angle \(x\) radians, so its area is \(\frac{x}{2}\). The outer triangle has perpendicular sides of lengths \(1\) and \(\tan x\), so its area is \(\frac{\tan x}{2}\).
The sector-area formula follows from comparing the sector's fraction of the full circle with its angle's fraction of a full turn. A full turn is \(2\pi\) radians, and a circle of radius \(r\) has area \(\pi r^2\). Thus a sector with angle \(x\) has area
The comparison of the three areas now gives an inequality. Since the radius is \(1\), it takes a particularly simple form.
From Areas to Bounds on the Quotient
Multiplying the area inequality by \(2\) gives \(\sin x<x<\tan x\). In this interval, \(\cos x>0\), and \(\tan x=\frac{\sin x}{\cos x}\). The right-hand inequality can therefore be rearranged without changing its direction:
Because \(x>0\), dividing through by \(x\) gives \(\cos x<\frac{\sin x}{x}\). Dividing \(\sin x<x\) by \(x>0\) gives the upper bound. Together, these bounds are
To use the Squeeze Theorem, the lower bound must approach \(1\) as \(x\) approaches zero. This can be seen without assuming the sine limit. The area comparison already gives \(0<\sin x<x\) for positive \(x\) near zero, so \(\sin x\to0\) as \(x\to0^+\). On this interval \(\cos x>0\), and the unit-circle identity \(\cos^2x+\sin^2x=1\) gives \(\cos x=\sqrt{1-\sin^2x}\). Continuity of the square-root function then gives \(\cos x\to1\).
Complete Derivation of the Two-Sided Limit
For \(0<x<\frac{\pi}{2}\), the area comparison gives \(\cos x<\frac{\sin x}{x}<1\).
As \(x\to0^+\), the unit-circle identity and the area bound give \(\cos x\to1\), while the constant upper bound also tends to \(1\).
Since both bounds tend to \(1\), \(\lim_{x\to0^+}\frac{\sin x}{x}=1\).
Sine is odd, so for \(x\ne0\), \(\frac{\sin(-x)}{-x}=\frac{\sin x}{x}\). The quotient is even; its left-hand limit therefore equals its right-hand limit.
The left- and right-hand limits both equal \(1\), so \(\lim_{x\to0}\frac{\sin x}{x}=1\).
This argument illustrates why the Squeeze Theorem is useful: the quotient does not need to be evaluated at zero. Instead, inequalities valid throughout a punctured interval determine how the quotient behaves as it approaches zero.
Worked Examples
Worked Example: Comparing the Areas at \(x=\frac{\pi}{6}\)
Take \(x=\frac{\pi}{6}\), which is between \(0\) and \(\frac{\pi}{2}\). The inner triangle has area
The sector has area
The outer triangle has area
Thus \(0.2500<0.2618<0.2887\), as the geometric comparison predicts. The corresponding quotient is \(\frac{\sin(\pi/6)}{\pi/6}=\frac{3}{\pi}\approx0.9549\), and the quotient bounds give \(\cos(\pi/6)=\frac{\sqrt{3}}{2}\approx0.8660<0.9549<1\). The computed quotient satisfies both bounds.
Worked Example: Using the Bounds at \(x=\frac{\pi}{3}\)
Since \(0<\frac{\pi}{3}<\frac{\pi}{2}\), the derived inequality applies. Its lower bound is
The quotient itself is
Therefore the bounds read \(\frac{1}{2}<\frac{3\sqrt{3}}{2\pi}<1\), or approximately \(0.5000<0.8270<1\). This checks the inequality at this angle; the limit proof depends on the bounds holding for every sufficiently small positive angle, not on this one example.
Worked Example: Checking the Effect of Measuring Angles in Degrees
Let \(t\) denote an angle measured in degrees, and let \(u=\frac{\pi t}{180}\) be its radian measure. The sine of the angle is \(\sin u\), so for \(t\ne0\),
As \(t\to0\), \(u=\frac{\pi t}{180}\to0\). The radian limit just proved gives \(\frac{\sin u}{u}\to1\), and hence
The result is not \(1\). The conversion factor is essential: the limit in the theorem uses the numerical value of the angle in radians.
Why Radians Are Required
The geometry depends on angle measure in two places. The area of a sector is \(\frac{1}{2}r^2x\) when \(x\) is measured in radians, and the coordinates of the point on the unit circle use \(\sin x\) and \(\cos x\) with that same radian measure. If the angle is instead recorded in degrees, the sector-area formula needs a conversion factor.
For example, the angle \(\frac{\pi}{3}\) radians is \(60\) degrees. The unit-circle sector has area \(\frac{1}{2}\cdot1^2\cdot\frac{\pi}{3}=\frac{\pi}{6}\). Substituting the degree number \(60\) into the radian formula would give \(\frac{60}{2}=30\), which is not the sector area. For degree measure \(d\), the correct sector area on a unit circle is \(\frac{d}{360}\pi\), not \(\frac{d}{2}\).
Common Mistake: Using the Sector Formula with Degrees
A common error is to write the sector area as \(\frac{x}{2}\) while treating \(x\) as a degree measure. For a \(30\)-degree sector in a unit circle, this incorrect calculation gives \(\frac{30}{2}=15\). The correct area is
The mistake is not a small rounding issue: it changes the relationship between angle and sector area on which the squeeze argument depends. Either use radians and the formula \(\frac{x}{2}\), or use degrees and the formula \(\frac{x}{360}\pi\); do not mix the two conventions.
Check Your Understanding
Use the area comparison and the reasoning in the derivation to answer these questions.
- For \(0<x<\frac{\pi}{2}\), what are the areas of the inner triangle, the sector, and the outer triangle?
- How does \(x<\tan x\) imply \(\cos x<\frac{\sin x}{x}\) on this interval?
- Why can the area bound \(\sin x<x\) help establish that \(\cos x\to1\) as \(x\to0^+\)?
- Why does the positive-angle argument also establish the limit from the left?
- What is \(\lim_{t\to0}\frac{\sin_{\mathrm{deg}}t}{t}\), and what conversion produces it?
