Squeeze theorem and trigonometric limits
Find a limit by trapping a function between simpler ones, then apply the idea to small angles.
The idea
An exact formula is not always the easiest route to a limit. The squeeze theorem determines what a function approaches by trapping its values between two simpler functions with the same limit.
Theorem
Squeeze theorem
Suppose, for all inputs sufficiently close to \(a\) except possibly \(a\) itself,
\[\ell(x)\le f(x)\le u(x).\]If \(\lim_{x\to a}\ell(x)=\lim_{x\to a}u(x)=L\), then
\[\lim_{x\to a}f(x)=L.\]The same statement holds for a one-sided approach.
Proof (optional)
Choose any small interval around \(L\). Since both bounding functions approach \(L\), they both stay inside that interval once \(x\) is sufficiently close to \(a\), with \(x\ne a\). The value \(f(x)\) lies between them, so it must stay inside the same interval. This is exactly what it means for \(f(x)\) to approach \(L\).
The bounds must close in on the same value; the trapped function can keep changing direction or oscillating.
Example 1
Oscillation inside a shrinking envelope
Let \(f(x)=x^2\sin(1/x)\) for \(x\ne0\). The sine bound gives
\[-x^2\le x^2\sin(1/x)\le x^2.\]Since \(\lim_{x\to0}(-x^2)=\lim_{x\to0}x^2=0\), the squeeze theorem gives
\[\lim_{x\to0}x^2\sin(1/x)=0.\]We do not use a product law with \(\sin(1/x)\): that factor has no limit at zero.
Two useful trigonometric limits
Theorem
Basic trigonometric limits
\[\begin{aligned} \lim_{t\to0}\frac{\sin t}{t}&=1,\\ \lim_{t\to0}\frac{1-\cos t}{t^2}&=\frac12. \end{aligned}\]Proof of the sine limit (optional)
For \(0<t<\pi/2\), the unit-circle arc \(AB\) has length \(t\). The chord is shorter, and is the hypotenuse of triangle \(ADB\), whose vertical side is \(BD=\sin t\). Thus
\[0\le\sin t\le AB\le t.\]Since \(\sin(-t)=-\sin t\), we obtain \(\lvert\sin t\rvert\le\lvert t\rvert\) near zero. Squeezing gives \(\lim_{t\to0}\sin t=0\); since cosine is positive near zero,
\[\lim_{t\to0}\cos t=\lim_{t\to0}\sqrt{1-\sin^2t}=1.\]For the other bound, triangle \(OAB\) lies inside the sector, inside triangle \(OAC\). Their areas give
\[\frac{\sin t}{2}\le\frac{t}{2}\le\frac{\tan t}{2} \quad\Longrightarrow\quad \cos t\le\frac{\sin t}{t}\le1.\]Here the sector area is half its angle on a unit circle; each triangle area is half its base times height. Squeezing yields the right-hand limit \(1\). The ratio is even, \(\sin(-t)/(-t)=\sin t/t\), so the left-hand limit is also \(1\).
Proof of the cosine limit (optional)
The preceding proof established \(\lim_{t\to0}\cos t=1\) as well as the sine ratio limit. For nearby \(t\ne0\), multiply by the conjugate:
\[\begin{aligned} \frac{1-\cos t}{t^2} &=\frac{1-\cos t}{t^2}\cdot\frac{1+\cos t}{1+\cos t}\\ &=\frac{1-\cos^2t}{t^2(1+\cos t)}\\ &=\frac{\sin^2t}{t^2(1+\cos t)} =\left(\frac{\sin t}{t}\right)^2\frac1{1+\cos t}. \end{aligned}\]The denominator approaches \(2\ne0\), so the limit laws give
\[\lim_{t\to0}\frac{1-\cos t}{t^2} =\left(\lim_{t\to0}\frac{\sin t}{t}\right)^2 \frac1{1+\lim_{t\to0}\cos t}=1^2\cdot\frac12=\frac12.\]The proofs also establish \(\lim_{t\to0}\sin t=0\) and \(\lim_{t\to0}\cos t=1\), which we can now use with the limit laws.
Worked examples
Example 2
Match the angle to the denominator
As \(x\to0\), the angle \(5x\to0\). For \(x\ne0\),
\[\frac{\sin(5x)}{2x}=\frac52\frac{\sin(5x)}{5x}.\]Therefore,
\[\lim_{x\to0}\frac{\sin(5x)}{2x} =\frac52\lim_{x\to0}\frac{\sin(5x)}{5x}=\frac52.\]Example 3
Tangent follows from sine and cosine
For nearby \(x\ne0\),
\[\frac{\tan x}{x}=\frac{\sin x}{x}\cdot\frac1{\cos x}.\]Using the two established limits and the quotient law,
\[\lim_{x\to0}\frac{\tan x}{x} =\left(\lim_{x\to0}\frac{\sin x}{x}\right)\frac1{\lim_{x\to0}\cos x}=1.\]Example 4
A smaller denominator changes the limit
The cosine limit has a squared denominator. For nearby \(x\ne0\),
\[\frac{1-\cos x}{x}=x\cdot\frac{1-\cos x}{x^2}.\]Consequently,
\[\lim_{x\to0}\frac{1-\cos x}{x} =\left(\lim_{x\to0}x\right) \left(\lim_{x\to0}\frac{1-\cos x}{x^2}\right)=0\cdot\frac12=0.\]The numerator is the same as in the featured limit, but the different denominator gives a different result.