The following notable limit is of fundamental importance as it frequently appears in limit calculations, derivative studies, and trigonometric function approximations.
We will provide two proofs: one based on a trigonometric identity, leveraging the relationship between sine and cosine, and another using the Taylor series expansion of the cosine function.
The notable limit we aim to prove is the following:
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \]
Trigonometric Identity
We use the identity:
\[ 1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right). \]
Rewriting the limit:
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2}. \]
Manipulating the expression:
\[ \frac{2\sin^2(x/2)}{x^2} = 2 \cdot \frac{\sin^2(x/2)}{(x/2)^2} \cdot \frac{(x/2)^2}{x^2}. \]
Using the well-known limit:
\[ \lim_{x \to 0} \frac{\sin(x/2)}{(x/2)} = 1, \]
we obtain:
\[ \lim_{x \to 0} 2 \cdot 1^2 \cdot \frac{1}{4} = \frac{1}{2}. \]
Taylor Series Expansion
The Taylor series expansion of \( \cos(x) \) is:
\[ \cos(x) = 1 - \frac{x^2}{2} + O(x^4). \]
From which we get:
\[ 1 - \cos(x) = \frac{x^2}{2} + O(x^4). \]
Now substituting into the limit:
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + O(x^4)}{x^2}. \]
Separating the terms:
\[ \lim_{x \to 0} \left(\frac{x^2}{2x^2} + \frac{O(x^4)}{x^2}\right) = \lim_{x \to 0} \left(\frac{1}{2} + O(x^2)\right). \]
Since \( O(x^2) \to 0 \) as \( x \to 0 \), the limit is:
\[ \frac{1}{2}. \]