What is this squeeze theorem? How it is useful? Where do we take this squeezing stance in the field of Calculus? Well, graphs of trigonometric functions are sketched by intuitive considerations because two concepts of Calculus, continuity & differentiation, are needed for a formal presentation of such graphs. Here in this article, we will discuss the continuity of trigonometric functions.

In our treatment of continuity, we apply the following limit:

\[\lim _{x\rightarrow 0}\dfrac{sint}{t}=1\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space (1)\]

Observe that f(t) does not exist when t=0 but does exist for all other values of t. To get an intuitive idea about the existence of the limit in equation (1) we first plot the graph of f(t).

Squeeze Theorem

Because f(0) does not exist, the graph has a “hole” on the y-axis. From the figure, we suspect that probably the limit in (1) does exist and is equal to 1. To examine the limit further, we compute on our calculator the function values given in flowing Table 1 and Table 2.

tf(t) = (sin t)/t
Table 1

tf(t) = (sin t)/t
Table 2

From the two tables we again suspect that if the limit in (1) exists, it may be equal to 1 (read epsilon-delta definition of limit), and in order to prove this we need the help of “Squeeze Theorem”. Though L’ Hopital’s rule is also a good remedy to sort out indeterminant forms like these, but indeed the mighty squeeze theorem is another weapon we have in proving the aforementioned equation (1) but also contributes to proving several other major theorems as well.

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Squeeze Theorem

Suppose that the functions f, g, and h are defined on some open interval “I” containing ‘a’ except possibly at ‘a’ itself, and that;

f(x) ≤ g(x) ≤ h(x)

for all ‘x’ is “I” for which x≠a. Also suppose that limx→a f(x) and limx→a h(x) both exist and are equal to L. Then limx→a g(x) exists and is equal to L.

Proof Of Squeeze Theorem

To prove that limx→a g(x) = L, we must show that for any ϵ > 0 there is a δ > 0 such that,

\[If\space\space\space\space\space 0 <\left| x-a\right| <\delta\space\space\space\space\space then\space\space\space\space\space \left| g\left( x\right) -L\right| <\varepsilon\]

We are given that,

limx→a f(x) = L              and              limx→a h(x) = L

and so any ϵ > 0 there is a δ > 0 such that;

\[if\space\space\space\space\space\space\space\space\space\space 0 <\left| x-a\right| <\delta _{1}\space\space\space\space\space \space\space\space\space\space then\space\space\space\space\space \space\space\space\space\space \left| f\left( x\right) -L\right| <\varepsilon\]
\[\Leftrightarrow\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <\left\| x-a\right\| <\delta _{1}\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space L-\varepsilon <f\left( x\right) <L+\varepsilon\]

And there is δ2 > 0 such that;

\[if\space\space\space\space\space\space\space\space\space\space 0 <\left| x-a\right| <\delta _{2}\space\space\space\space\space \space\space\space\space\space then\space\space\space\space\space \space\space\space\space\space \left| h\left( x\right) -L\right| <\varepsilon\]
\[\Leftrightarrow\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <\left\| x-a\right\| <\delta _{2}\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space L-\varepsilon <h\left( x\right) <L+\varepsilon\]

Let δ = min (δ1 , δ2) and so δ ≤ δ1 and δ ≤ δ2. Therefore, in the light of the above inequalities;

\[if\space\space\space\space\space\space\space\space\space\space 0 <\left| x-a\right| <\delta\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space L-\varepsilon <f\left( x\right)\]


\[if\space\space\space\space\space\space\space\space\space\space 0 <\left| x-a\right| <\delta\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space h\left( x\right) <L+\varepsilon\]

We are given that;

f(x) ≤ g(x) ≤ h(x)

This reveals that,

\[\Leftrightarrow\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <\left\| x-a\right\| <\delta\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space L-\varepsilon < f(x) ≤ g(x) ≤ h(x)< L+\varepsilon\]


\[\Leftrightarrow\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <\left\| x-a\right\| <\delta\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space L-\varepsilon <g\left( x\right) <L+\varepsilon\]
\[\Leftrightarrow\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <\left| x-a\right| <\delta\space\space\space\space\space\space\space\space\space\space then\space\space\space\space\space\space\space\space\space\space \left| g\left( x\right) -L\right| <\varepsilon\]

which yields;

\[\lim _{x\rightarrow a}g\left( x\right) =1\]

hence proved.

Now let’s try to understand some important theorems of trig-functions continuity with the help of the squeeze theorem.

Theorem: limt→0 (sint)/t =1


First assume that 0<t<π/2, the following figure;

Continuity of trig functions

Shows a unit circle x2 + y2 = 1 and the shaded sector BOP, where B is the point (1,0) and P is the point (cos t, sin t). The area of a circular sector of radius ‘r’ and central angle of radian measure t is determined by ½ r2t; so, if S square units is the area of sector BOP,

S = ½ t                         (2)

Consider now ∆BOP, and let K1 square units be the area of this triangle. Because;

\[K_{1}=\dfrac{1}{2}\left| \overline{AP}\right| \cdot \left| \overline{OB}\right|\space\space\space\space\space\space\space\space\space\space , \space\space\space\space\space\space\space\space\space\space \left| \overline{AP}\right| =\sin t\space\space\space\space\space\space\space\space\space\space and\space\space\space\space\space\space\space\space\space\space \left| \overline{OB}\right| =1\]

We have;

\[K_{1}=\dfrac{1}{2}\sin t\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space (3)\]

If Ksquare units is the area of the right triangle BOT, where T is the point (1, tan t), then;

\[K_{2}=\dfrac{1}{2}\tan t\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space (4)\]

From the above figure, observe that;

K1 < S < K2

Substituting from (2), (3), and (4) into the inequality;

\[\dfrac{1}{2}\sin t <\dfrac{1}{2}t <\dfrac{1}{2}\tan t\]

Multiplying each member of this inequality by 2/sin t, which is positive because 0<t<π/2, we have;

\[1 <\dfrac{t}{\sin t} <\dfrac{1}{\cos t}\space\space\space\space\space\space\space\space\space\space ( because\space\space\space\space\space\space \dfrac{\tan t}{\sin t}= \dfrac{1}{\cos t})\]

Taking the reciprocal of each member of this inequality which reverses the direction of the inequality signs;

\[\cos t <\dfrac{\sin t}{t} <1\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \left( 5\right)\]

From the right-hand inequality in the above;

sin t < t                    (6)

and from the half-measure identity in trigonometry,

\[\dfrac{1-\cos t}{2}=\sin ^{2}\dfrac{t}{2}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \left( 7\right)\]

Replacing t by ½ t in inequality (6) and squaring;

\[\sin ^{2}\dfrac{t}{2} <\dfrac{1}{4}t^{2}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \left( 8\right)\]

Thus from (7) & (8), we get;

\[\dfrac{1-\cos t}{2} <\dfrac{t^{2}}{4}\]
\[\Leftrightarrow\space\space\space\space\space\space 1-\dfrac{1}{2}t^{2} <\cos t\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \left( 9\right)\]

From (5) and (12) and because 0<t<π/2,

\[1-\dfrac{t^{2}}{2} <\dfrac{\sin t}{t} <1\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space 0 <t <\dfrac{\pi }{2}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \left( 10\right)\]

If -π/2 < t < 0, then 0 < -t < π/2; so from (10);

\[1-\dfrac{1}{2}\left( -t\right) ^{2} <\dfrac{\sin \left( -t\right) }{-t} <1\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space -\dfrac{\pi }{2} <t <0\space\space\space\space\space\space\space\space\space\space (11)\]


sin (-t) = -sin t

thus equation (11) can be written as;

\[1-\dfrac{t^{2}}{2} <\dfrac{\sin t}{t} <1\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space -\dfrac{\pi }{2} <t <0\space\space\space\space\space\space\space\space\space\space\space\space \left( 12\right)\]

From (10) and (12) we conclude that;

\[1-\dfrac{t^{2}}{2} <\dfrac{\sin t}{t} <1\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space \dfrac{-\pi }{2} <t <\dfrac{\pi }{2}\space\space\space\space\space\space\space\space\space\space and\space\space\space\space\space\space\space\space\space\space t\neq 0\space\space\space\space\space\space\space\space\space\space \left( 13\right)\]


\[\lim _{t\rightarrow 0}\left( 1-\dfrac{t^{2}}{2}\right) =1\space\space\space\space\space\space\space\space\space\space and\space\space\space\space\space\space\space\space\space\space \lim _{t\rightarrow 0}1=1\]

It follows from (13) and squeeze theorem that;

\[\lim _{t\rightarrow 0}\dfrac{\sin t}{t}=1\]

hence proved.

Theorem: limt→0 (1-cos t)/t =0


\[\lim _{t\rightarrow 0}\dfrac{1-\cos t}{t}=\lim _{t\rightarrow 0}\dfrac{\left( 1-\cos t\right) \left( 1+\cos t\right) }{t\left( 1+\cos t\right) }\]
\[\begin{aligned}\Leftrightarrow \lim _{t\rightarrow 0}\dfrac{1-\cos ^{2}t}{t\left( 1+\cos t\right) }\\ \Leftrightarrow \lim _{t\rightarrow 0}\dfrac{\sin ^{2}t}{t\left( 1+\cos t\right) }\\\Leftrightarrow \lim _{t\rightarrow 0}\dfrac{\sin t}{t}\cdot \lim _{t\rightarrow 0}\dfrac{sin t}{1+\cos t}\end{aligned}\]


\[\lim _{t\rightarrow 0}\dfrac{\sin t}{t}=1\]

And because the sine and cosine functions are continuous at ‘0’ it follows that;

\[\lim _{t\rightarrow 0}\dfrac{\sin t}{1+\cos t}=\dfrac{0}{1+1}=0\]


\[\lim _{t\rightarrow 0}\dfrac{1-\cos t}{t}=1\times 0=0\]

hence proved.

Use Squeeze Theorem To Prove That limx→0 |x sin(1/x)| =0. Support this fact graphically


Because -1 ≤ sin t ≤ 1 for all t, then;

\[0\leq \left| x\cdot \sin \dfrac{1}{x}\right| \leq 1\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space x\neq 0\]

Therefore, if x ≠ 0,

\[\left| x\cdot sin\dfrac{1}{x}\right| =\left| x\right| \cdot \left| \sin \dfrac{1}{x}\right| \leq \left| x\right|\]


\[0\leq \left| x\cdot \sin \dfrac{1}{x}\right| \leq \left| x\right|\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space x\neq 0\]

Because limx→0 0 = 0 and limx→0 |x| = 0, it follows from above inequality and the squeeze theorem that;

\[\lim _{x\rightarrow 0}\left| x\cdot \sin \dfrac{1}{x}\right| =0\]

The graph of the function having values |x. sin 1/x| plotted within the domain [-1, 1],

example of squeeze theorem

observe the unusual oscillating behavior of the function when -0.32 ≤ x ≤ 0.32. The graph supports the fact that the limit is 0.


It’s not like each and every single topic of mathematics has its significance in everyday life. We can see squeeze theorem apparently has no contribution in real life but its involvement in proving the continuity of typical and absurd-looking functions reveals its deep role in Calculus, and we know how differential and integral Calculus has influenced multifarious fields of science and technology. So that’s how the squeeze theorem indirectly puts its impact on day-to-day life.

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