The following limit is of fundamental importance in mathematical analysis. We will provide two proofs: one based on the Squeeze Theorem (also known as the Sandwich Theorem), which exploits a geometric comparison, and another using the Taylor Series, which employs the series expansion of the sine function. The limit we intend to prove is:
\[\lim_{x \to 0} \frac{\sin(x)}{x}\]
Proof using the Squeeze Theorem
To prove the limit, we use the Squeeze Theorem (also known as the Sandwich Theorem). Consider the figure below:
We consider a circular sector with radius \(1\) and angle \(x\) in radians. As shown in the figure, we can compare three areas:
- The triangle \( \triangle OBD \), with base \( \cos x \) and height \( \sin x \).
- The circular sector \( OBD \), whose area is proportional to \( x \) (since the area of a sector is given by \( \displaystyle \frac{1}{2}r^2 x \) and, for a unit radius, it becomes \( \displaystyle \frac{x}{2} \)).
- The triangle \( \triangle OBC \), with base \(1\) and height \( \tan x \).
These three regions satisfy the relation:
\[ \text{Area}(\triangle OBD) < \text{Area}(OBC) < \text{Area}(\triangle OBC) \]
Expressing the areas explicitly, we obtain:
\[ \frac{1}{2}\cos x \sin x < \frac{1}{2}x < \frac{1}{2}\tan x \]
Multiplying every part by \( \frac{2}{\sin x} \), we have:
\[ \cos x < \frac{x}{\sin x} < \frac{1}{\cos x} \]
Since we know that:
\[ \lim_{x\to 0} \cos x = 1, \quad \lim_{x\to 0} \frac{1}{\cos x} = 1, \]
by the Squeeze Theorem we can conclude that:
\[ \lim_{x\to 0} \frac{x}{\sin x} = \lim_{x\to 0} \frac{\sin x}{x} = 1. \]
Proof using the Taylor Series
Another rigorous method is to use the Taylor series expansion of \( \sin(x) \) around \( x = 0 \):
\[ \sin(x) = x - \frac{x^3}{3!} + \mathcal{O}(x^5). \]
Dividing everything by \( x \):
\[ \frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \mathcal{O}(x^4). \]
Letting \( x \to 0 \), the higher order terms vanish, and we obtain:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1. \]
We have demonstrated that the limit
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]
using both the Squeeze Theorem and the Taylor Series. This result is fundamental in analysis and has numerous applications.
Example
Notice that in the sine function the argument appears as \( ax \) instead of \( x \). To adapt it to the fundamental limit, we multiply and divide by \( a \):
\[ \frac{\sin(ax)}{x} = a \cdot \frac{\sin(ax)}{ax}. \]
Now, the term \( \displaystyle \frac{\sin(ax)}{ax} \) has the same structure as the fundamental limit.
Applying the Fundamental Limit
Since we know that:
\[ \lim_{y \to 0} \frac{\sin y}{y} = 1, \]
we can substitute \( y = ax \), which tends to \( 0 \) as \( x \to 0 \), obtaining:
\[ \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1. \]
Multiplying by \( a \), we obtain the final result:
\[ \lim_{x \to 0} \frac{\sin(ax)}{x} = a. \]
Numerical Example
Consider a specific case: if \( a = 3 \), then the limit becomes:
\[ \lim_{x \to 0} \frac{\sin(3x)}{x}. \]
Applying the derived formula, we obtain:
\[ \lim_{x \to 0} \frac{\sin(3x)}{x} = 3. \]