Squeeze Theorem & Continuity Of Trig-Functions|| Proofs & Examples

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.

Table of Content

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).

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.

t	f(t) = (sin t)/t
1.0	0.84147
0.9	0.87036
0.8	0.89670
0.7	0.92031
0.6	0.94107
0.5	0.95885
0.4	0.97355
0.3	0.98507
0.2	0.99335
0.1	0.99833
0.01	0.99998

Table 1

t	f(t) = (sin t)/t
-1.0	0.84147
-0.9	0.87036
-0.8	0.89670
-0.7	0.92031
-0.6	0.94107
-0.5	0.95885
-0.4	0.97355
-0.3	0.98507
-0.2	0.99335
-0.1	0.99833
-0.01	0.99998

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.

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 lim_x→a f(x) and lim_x→ah(x) both exist and are equal to L. Then lim_x→a g(x) exists and is equal to L.

Proof Of Squeeze Theorem

To prove that lim_x→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,

lim_x_→af(x) = L and lim_x_→ah(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;

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

and;

We are given that;

f(x) ≤ g(x) ≤ h(x)

This reveals that,

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

Therefore;

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

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

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: lim_t→0 (sint)/t =1

Proof:

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

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 K₂square 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);

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

But;

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;

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

Because;

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

It follows from (13) and squeeze theorem that;

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

hence proved.

Theorem: lim_t→0 (1-cos t)/t =0

Proof:

\[\lim _{t\rightarrow 0}\dfrac{1-\cos t}{t}=\lim _{t\rightarrow 0}\dfrac{\left( 1-\cos t\right) \left( 1+\cos t\right) }{t\left( 1+\cos t\right) }\]

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

As;

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

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

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

Therefore;

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

hence proved.

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

Proof:

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,

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

Hence;

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

Because lim_x→0 0 = 0 and lim_x→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],

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

Conclusion

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.

Q1) What is the squeeze theorem?
The squeeze theorem, also known as the sandwich theorem, is a theorem in calculus that states that if a function f(x) is sandwiched between two other functions g(x) and h(x) such that g(x)≤f(x)≤h(x) for all x in a given interval, and the limits of g(x) and h(x) as x approaches a certain value c are both equal to L, then the limit of f(x) as x approaches c is also equal to L.

Q2) Why do we use the squeeze theorem?
The squeeze theorem is used to find the limits of functions that are difficult to evaluate directly. By sandwiching the function between two other functions whose limits are known, we can use the squeeze theorem to conclude that the limit of the original function is also equal to the known limit.

Q3) What is a squeeze theorem calculator?

A squeeze theorem calculator is a tool that can be used to find the limits of functions using the squeeze theorem. The calculator typically takes as input the function f(x), the two functions g(x) and h(x), and the point c where we are taking the limit. The calculator then checks to see if the conditions for the squeeze theorem are met. If they are, the calculator will output the limit of f(x) as x approaches c.

Q4) How do I use a squeeze theorem calculator?

To use a squeeze theorem calculator, you will need to know the function f(x), the two functions g(x) and h(x), and the point c where you are taking the limit. Once you have this information, you can input it into the calculator. The calculator will then output the limit of f(x) as x approaches c.

Q5) What are the limitations of squeeze theorem calculators?

Squeeze theorem calculators are only as good as the functions that they are given. If the functions are not defined or continuous on the interval (a,b), then the calculator will not be able to find the limit. Additionally, if the functions g(x) and h(x) do not have the same limit as x approaches c, then the calculator will not be able to find the limit of f(x) either.

Q6) Where can I find a squeeze theorem calculator?

There are many squeeze theorem calculators available online. Some of these calculators are free to use, while others require a subscription. You can find a list of squeeze theorem calculators by searching for “squeeze theorem calculator” on the internet.

Q7) Does the continuity of trigonometric functions affect the squeeze theorem?

The continuity of the trigonometric functions does not affect the squeeze theorem. The squeeze theorem can be used to find the limits of trigonometric functions, even if the functions are not continuous. However, the continuity of the trigonometric functions can make it easier to find the functions that can be used to sandwich the function of interest. For example, in the previous example, we knew that the functions g(x)=sinx and h(x)=x were both continuous at x=0, so we could be sure that they would work as the sandwiching functions.

Q8) What are some other ways to find the limits of trigonometric functions?

There are many other ways to find the limits of trigonometric functions. Some of these methods include: