Why This Limit Matters
The quotient \(\frac{\sin x}{x}\) is not defined at \(x=0\), so direct substitution cannot determine its limit there. Yet its values near zero approach a simple number. Establishing that number by a precise argument is important: the result is used throughout calculus, including in later work with trigonometric limits and derivatives.
The key is to compare the quotient with quantities whose behavior near zero can be controlled. Geometry supplies the comparisons, and the Squeeze Theorem turns them into a limit.
The Geometric Bounds
On a unit circle, take an angle \(t\) with \(0<t<\frac{\pi}{2}\). The area of the triangle formed by the two radii and the chord is \(\frac{1}{2}\sin t\). The sector has area \(\frac{t}{2}\), because the angle is measured in radians. The triangle formed by the two radii and the tangent line has area \(\frac{1}{2}\tan t\). The first triangle lies inside the sector, and the sector lies inside the tangent triangle. Comparing their areas gives the geometric bounds below.
Multiplying by \(2\) gives \(\sin t<t<\tan t\). Since \(\tan t=\frac{\sin t}{\cos t}\) and \(\cos t>0\) in this range, these inequalities can be rewritten as bounds for the quotient.
For the lower bound, \(t<\frac{\sin t}{\cos t}\) implies \(t\cos t<\sin t\). Dividing by \(t>0\) gives \(\cos t<\frac{\sin t}{t}\). The upper bound follows by dividing \(\sin t<t\) by \(t>0\).
Worked Examples with the Bounds
Worked Example: Bounding the Quotient at \(\frac{\pi}{6}\)
Use the geometric bounds to place \(\frac{\sin(\pi/6)}{\pi/6}\) between two exact values.
First, \(\frac{\pi}{6}\) is positive and less than \(\frac{\pi}{2}\), so the quotient bounds apply. The lower bound is \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\). The upper bound is \(1\).
Now compute the quotient using \(\sin(\frac{\pi}{6})=\frac{1}{2}\):
Thus \(\frac{\sqrt{3}}{2}<\frac{3}{\pi}<1\). This agrees with the geometric bounds: the angle is in the required interval, and the computed quotient lies between its cosine and \(1\).
Worked Example: Bounding the Quotient for a Negative Angle
Find a corresponding bound for \(\frac{\sin(-\pi/6)}{-\pi/6}\).
Use \(\sin(-t)=-\sin t\). Then
The positive-angle bounds from the previous example therefore give
As a check, substituting the exact sine value directly gives \(\frac{(-1/2)}{(-\pi/6)}=\frac{3}{\pi}\), the same quotient and bounds. The negative signs cancel; the quotient is positive.
Worked Example: Estimating How Far the Quotient Is from 1
For \(t=\frac{1}{2}\), use the bounds to estimate \(1-\frac{\sin t}{t}\).
Because \(0<\frac{1}{2}<\frac{\pi}{2}\), the quotient satisfies
For any \(0<t<1\), the geometric bound \(\sin t<t\) gives \(\sin^2t<t^2\). Since \(\cos t>0\) in this range and \(\sin^2t+\cos^2t=1\), it follows that
At \(t=\frac{1}{2}\), this gives \(\cos(\frac{1}{2})>\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\). Therefore,
Subtracting these bounds from \(1\), with the inequality directions reversed, yields
The check is that the quotient lies below \(1\), so its difference from \(1\) is positive, and that the displayed upper bound is positive because \(\frac{\sqrt{3}}{2}<1\).
Getting a Lower Bound That Also Approaches 1
The quotient is bounded above by \(1\). To use the Squeeze Theorem, we also need a lower bound that approaches \(1\). For \(0<t<1\), the geometric inequality \(\sin t<t\) and the identity \(\sin^2t+\cos^2t=1\) provide such a bound.
For \(0<t<1\), the geometric inequality applies and gives \(0<\sin t<t\). Squaring preserves the inequality because both sides are positive, so \(\sin^2t<t^2\).
The identity \(\cos^2t=1-\sin^2t\) gives \(\cos^2t>1-t^2\). Both \(\cos t\) and \(\sqrt{1-t^2}\) are positive for \(0<t<1\). Taking positive square roots therefore gives \(\cos t>\sqrt{1-t^2}\).
The geometric quotient bound gives \(\frac{\sin t}{t}>\cos t\). Combining the inequalities yields \(\sqrt{1-t^2}<\frac{\sin t}{t}<1\).
As \(t\) approaches \(0\) through positive values, \(1-t^2\) approaches \(1\). The root limit law, with the radicand positive near \(0\), gives \(\sqrt{1-t^2}\to1\). The constant upper bound \(1\) also approaches \(1\). This provides the bounds needed to squeeze the positive-side quotient to \(1\).
Proving the Limit on Both Sides of Zero
The geometric bounds above are stated for positive angles. To handle negative \(x\), set \(t=|x|\), the nonnegative distance from \(x\) to zero. For \(x\ne0\), the oddness identity \(\sin(-t)=-\sin t\) shows that \(\frac{\sin x}{x}=\frac{\sin|x|}{|x|}\). Thus the quotient has the same value at \(x\) and at \(-x\).
Whenever \(0<|x|<1\), the positive-angle bounds applied to \(t=|x|\) and the lower-bound argument above give
The lower bound approaches \(1\) as \(x\to0\), since \(|x|^2\to0\) and the root limit law applies. The upper bound is the constant \(1\). Therefore the Squeeze Theorem proves the desired limit.
This argument checks the conditions of the squeeze directly: the inequalities hold throughout a punctured neighborhood of zero, the middle expression is defined there, and both bounding expressions have the same limit. The value of the quotient at \(x=0\) is not needed and is not defined.
Common Pitfalls
One error is to use the positive-angle inequality \(\sin t<t\) as though it directly gave \(\sin x<x\) for negative \(x\). For example, at \(x=-\frac{\pi}{6}\), \(\sin x=-\frac{1}{2}\), while \(x=-\frac{\pi}{6}\); in fact \(-\frac{1}{2}>-\frac{\pi}{6}\), not \(\sin x<x\). The correction is to apply the positive-angle bounds to \(|x|\), and then use \(\frac{\sin x}{x}=\frac{\sin|x|}{|x|}\).
A second error is to treat the angle as measured in degrees while keeping the same quotient. For instance, \(\sin(30^\circ)=\frac{1}{2}\), so dividing by the numerical degree measure \(30\) gives \(\frac{1}{60}\), not a value near \(1\). The geometric sector area formula used above is \(\frac{t}{2}\) only when \(t\) is measured in radians. The quotient limit in this form is a radian result.
Finally, an inequality alone does not prove a limit. The bounds must be valid near the point, and both bounding expressions must approach the same number. Here the lower bound \(\sqrt{1-|x|^2}\) and upper bound \(1\) both approach \(1\), which is precisely why squeezing works.
Why the Result Is Useful
The limit identifies the local behavior of sine near zero: for small radian angles, \(\sin x\) is close to \(x\), in the precise sense that their quotient approaches \(1\). This is not a claim that \(\sin x=x\) for nonzero \(x\); rather, the relative difference becomes small as the angle approaches zero. The geometric proof makes clear both the role of radians and the reason the quotient is controlled from above and below.
Check Your Understanding
Use the geometric bounds and the squeeze argument to answer these questions.
- For which positive angles does the geometric inequality \(\sin t<t<\tan t\) apply?
- How does \(\sin t<t<\tan t\) lead to \(\cos t<\frac{\sin t}{t}<1\) when \(t>0\)?
- Why does the quotient \(\frac{\sin x}{x}\) have the same value at \(x\) and \(-x\), when \(x\ne0\)?
- How does \(\sin^2t<t^2\) imply \(\cos t>\sqrt{1-t^2}\) for \(0<t<1\)?
- Why must the angle be measured in radians for this geometric proof and limit statement?
