Limit of (1 - cos x) over x squared as x approaches 0
This standard limit characterizes the behavior of the cosine function as \( x \) approaches zero. \[
\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
This result shows that, in a neighborhood of zero, the expression \( 1 - \cos x \) is asymptotically equivalent to a quantity proportional to \( x^2 \). More precisely, for small values of \( x \), we have the approximation \( 1 - \cos x \approx \frac{x^2}{2} \).
This limit is frequently used to simplify expressions, evaluate more involved limits, and analyze the local behavior of trigonometric functions near the origin.
Proof
Consider the limit
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]
This is an indeterminate form, since as \( x \to 0 \) we obtain:
\[ \lim_{x \to 0} \frac{1 - \cos 0}{0^2} = \frac{0}{0} \]
To resolve it, multiply and divide by \( 1 + \cos x \):
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} \]
\[ \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)} \]
\[ \lim_{x \to 0} \frac{1 - \cos^2 x}{x^2(1 + \cos x)} \]
Using the identity \( 1 - \cos^2 x = \sin^2 x \), which follows from the Pythagorean identity, we obtain:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2(1 + \cos x)} \]
Now rewrite the expression as a product:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{1}{1 + \cos x} \]
At this point, apply standard limits and direct substitution:
- This is the standard sine limit, whose value is known
\( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) - The second limit is evaluated by direct substitution
\( \lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{1 + \cos 0} = \frac{1}{2} \)
Therefore, as \( x \to 0 \), the limit becomes:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{1}{1 + \cos x} = 1^2 \cdot \frac{1}{2} = \frac{1}{2} \]
Hence, the original limit is equal to \( \frac{1}{2} \):
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
As required.
Alternative proof
An alternative derivation can be obtained as follows.
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]
Use the double-angle identity:
\[ 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right) \]
Substitute into the limit:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2} \]
Rewrite \( x^2 \) as \( 4 \left( \frac{x}{2} \right)^2 \):
\[ \lim_{x \to 0} \frac{2 \sin^2 \left( \frac{x}{2} \right)}{4 \left( \frac{x}{2} \right)^2} \]
\[ \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin \left( \frac{x}{2} \right)}{\frac{x}{2}} \right)^2 \]
This isolates the standard sine limit:
\[ \lim_{x \to 0} \frac{\sin \left( \frac{x}{2} \right)}{\frac{x}{2}} = 1 \]
Therefore, the limit is equal to \( \frac{1}{2} \):
\[ \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin \left( \frac{x}{2} \right)}{\frac{x}{2}} \right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2} \]
Hence, the original limit is again equal to \( \frac{1}{2} \):
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
As required.
And so on.
