The following limit is of fundamental importance as it frequently appears in the study of derivatives, limits, and approximations of trigonometric functions.
We will provide two proofs: one based on the Comparison Theorem (also known as the Squeeze Theorem), which leverages a geometric comparison, and another using the Taylor Series, which employs the series expansion of the cosine function.
The limit we aim to prove is the following:
\[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1 \]
Proof Using the Comparison Theorem
Consider the unit circle and an angle \( x \) measured in radians. It is well known that:
\[ \sin(x) < x < \tan(x), \quad \text{for } x \in (0, \frac{\pi}{2}). \]
Dividing by \( \sin(x) \):
\[ 1 < \frac{x}{\sin(x)} < \frac{\tan(x)}{\sin(x)}. \]
We can rewrite the fraction as:
\[ \frac{\tan(x)}{x} = \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)}. \]
Since it is well known that:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1, \quad \text{and} \quad \cos(x) \to 1, \]
it follows that the reciprocal \( \frac{1}{\cos(x)} \) also tends to \( 1 \), and we conclude:
\[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1. \]
Proof Using Taylor Series
We use the Taylor series expansions of the sine and cosine functions:
\[ \sin(x) = x - \frac{x^3}{6} + O(x^5), \] \[ \cos(x) = 1 - \frac{x^2}{2} + O(x^4). \]
We compute the reciprocal of \( \cos(x) \) using the series expansion of \( \frac{1}{1 - u} \):
\[ \frac{1}{\cos(x)} = 1 + \frac{x^2}{2} + O(x^4). \]
Now, we multiply by \( \sin(x) \):
\[ \tan(x) = (x - \frac{x^3}{6} + O(x^5)) \cdot (1 + \frac{x^2}{2} + O(x^4)). \]
Expanding the leading terms:
\[ \tan(x) = x + \frac{x^3}{2} - \frac{x^3}{6} + O(x^5). \]
Since
\[ \frac{x^3}{2} - \frac{x^3}{6} = \frac{x^3}{3}, \]
we obtain
\[ \tan(x) = x + \frac{x^3}{3} + O(x^5). \]
Dividing by \( x \):
\[ \frac{\tan(x)}{x} = 1 + \frac{x^2}{3} + O(x^4). \]
Since the term \( \frac{x^2}{3} + O(x^4) \) tends to zero as \( x \to 0 \), it follows that:
\[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1. \]