Proof of Lim x→0 sinx/x=1

In this discussion, we are going to prove

limθ→0sinθθ=1

This is an important limit in calculus, as it will help you find the limits of other trig ratios.

Want to see some examples of limits of trig ratios?

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Here is an outline of this discussion:

🔹Using the Unit Circle

🔹Using Inequalities

🔹Finding the Area

🔹Finding the Limits

🔹Picture of the Graph

It is assumed that you understand the unit circle and right triangle trigonometry.

We begin with what we already know. Recall, the unit circle!

Unit circle diagram used to begin the geometric proof that the limit of sin x divided by x as x approaches zero equals 1.

🔹Using the Unit Circle

Draw a positive angle in quadrant 1 and call the angle θ.

Notice we have a Sector.

Unit circle with a positive angle theta in the first quadrant forming a sector for the geometric proof of the sin x over x limit.

Draw a line from the terminal point of the angle θ to the terminal point of the angle 0.

Notice we have a triangle and a sector.

Since we are on the unit circle, we know that the radius is 1.

We also know that the distance from the center of the circle to any point on the circle is 1.

Now draw a line straight up from the terminal point of angle 0 and extend the line from the terminal point of the angle θ until the two lines meet.

Now we have a bigger triangle.

Now, let’s pull out the smaller triangle, the sector, and the bigger triangle and draw them separately.

Draw a dotted line to indicate the height of the smaller triangle.

Let the height of the smaller triangle equal h1.

Let the height of the bigger triangle equal h2.

Note that the dotted line creates a 90-degree angle for the smaller triangle and the bigger triangle also has a 90-degree angle.

We know that the base of the two triangles is equal to 1 and the radius of the sector is equal to 1.

Let

b = the base of the two triangles

and

r = the radius of the sector.

We can find h1 and h2 using Right-Triangle Trigonometry.

For the smaller triangle:

\sin θ = \frac{opposite}{hypotenuse}

\sin θ = \frac{h₁}{1}

\sin θ = h₁

For the larger triangle:

\tan θ = \frac{opposite}{adjacent}

\tan θ = \frac{h₂}{1}

\tan θ = h₂

🔹Using Inequalities

Compare the area of the smaller triangle, the sector, and the larger triangle.

We know that:

The area of the smaller triangle < the area of the sector < the area of the bigger triangle.

🔹Finding the Area

Area of the smaller triangle	<	Area of the sector	<	Area of the bigger triangle
$\frac{1}{2}$ base · height	<	$\frac{1}{2}$ radius² · θ in radians	<	$\frac{1}{2}$ base · height
$\frac{1}{2}$ b · h₁	<	$\frac{1}{2}$ r² · θ	<	$\frac{1}{2}$ b · h₂
$\frac{1}{2}$ (1) sin θ	<	$\frac{1}{2}$ (1)² θ	<	$\frac{1}{2}$ (1) tan θ
$\frac{\sin θ}{2}$	<	$\frac{θ}{2}$	<	$\frac{\tan θ}{2}$

Note that, any time you are working with a sector, the angle must be in radians.

Radians help us relate linear and angular measurements.

For example, we know that the area of a circle is

A=πr2,

which is an angle given in radians.

A sector is a segment of a circle. Thus, the area of a sector is a fraction of the area of a circle.

So, the area of a sector =

π r^{2} \cdot \frac{θ}{2 π} = \frac{1}{2} r^{2} θ

That was a little off subject, so let us continue where we left off.

We have the following inequality:

\frac{\sin θ}{2} < \frac{θ}{2} < \frac{\tan θ}{2}

Now the above inequality is true for positive θ in the first quadrant.

This means that as θ gets closer to 0, we are approaching 0 from the right.

However, we need to know what happens as we approach 0 from the left because, for the limit as θ approaches 0 to exist, the right-hand limit must equal the left-hand limit.

Thus, for the inequality to hold true for negative θ in the fourth quadrant, we take the absolute value of the inequality.

\frac{| \sin θ |}{2} < \frac{| θ |}{2} < \frac{| \tan θ |}{2}

We can do this because, if I were to draw the triangles and sector in the fourth quadrant using negative θ, there would be symmetry about the horizontal axis.

Thus, the distances for negative θ will be the same as the distances for positive θ.

Although we get negative values for sin(θ) and tan(θ) in the fourth quadrant, area is always positive.

Thus, taking the absolute value of the inequality holds true for positive and negative θ.

Thus, we have,

\begin{array}{ccccc} \frac{1}{2} {base}_{1} {height}_{1} & < & \frac{1}{2} {radius}^{2} θ & < & \frac{1}{2} {base}_{2} {height}_{2} \\ \frac{1}{2} b_{1} h_{1} & < & \frac{1}{2} r^{2} θ & < & \frac{1}{2} b_{2} h_{2} \\ \frac{1}{2} (1) \sin θ & < & \frac{1}{2} (^{1})^{2} θ & < & \frac{1}{2} (1) \tan θ \\ \frac{\sin θ}{2} & < & \frac{θ}{2} & < & \frac{\tan θ}{2} \end{array}

For −θ, take the absolute value.

\frac{| \sin θ |}{2} < \frac{| θ |}{2} < \frac{| \tan θ |}{2}

Multiply through by 2 and rewrite tan θ as sin θ / cos θ.

| \sin θ | < | θ | < | \frac{\sin θ}{\cos θ} |

Dividing through by sin θ

1 < | \frac{θ}{\sin θ} | < \frac{1}{\cos θ}

Taking reciprocals, we reverse the inequalities.

1 > | \frac{\sin θ}{θ} | > | \cos θ |

Remove the absolute values since sin(−θ)/(−θ) = (−sin θ)/(−θ) = sin θ/θ and cos(−θ) = cos θ.

1 > \frac{\sin θ}{θ} > \cos θ

Now, lim_θ→0 1 = 1 and lim_θ→0 cos θ = cos(0) = 1.

\lim_{θ \to 0} \frac{\sin θ}{θ} = 1

By the Squeeze Theorem.

This completes the proof.

Hence, proved.

🔹Picture of the Graph

y = 1 (in blue)

y = sinx/ x (in red)

y = cosx (in purple)

We see that, as x approaches 0, all functions converge to 1 as x approaches 0.

I hope my Blog helps you.

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Proof of Lim x→0 sinx/x=1

🔹Using the Unit Circle

🔹Using Inequalities

🔹Finding the Area

🔹Picture of the Graph

More on Limits

🎥Suggested Playlist: Limits

✨Extra Resources✨

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