Limit of (1 - cos x) / x as x approaches 0
This fundamental limit states that the ratio between \(1 - \cos x\) and \(x\) approaches zero as \(x\) approaches zero. $$ \lim_{x \to 0} \frac{1-\cos x}{x} = 0 $$
This result plays an important role in the study of limits involving trigonometric functions and is frequently used to simplify the evaluation of more complicated limits.
The reason is that, for very small values of \(x\), the function \( \cos x \) is extremely close to \(1\). As a consequence, the difference \(1 - \cos x\) becomes negligible compared with \(x\).
Proof
This limit has the indeterminate form
$$ \lim_{x \to 0} \frac{1-\cos x}{x} = \frac{0}{0} $$
To evaluate it, multiply and divide the expression by \(1 + \cos x\).
$$ \lim_{x \to 0} \frac{1-\cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} $$
$$ \lim_{x \to 0} \frac{1-\cos^2 x}{x \cdot (1 + \cos x)} $$
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we obtain \(1 - \cos^2 x = \sin^2 x\).
$$ \lim_{x \to 0} \frac{\sin^2 x}{x \cdot (1 + \cos x)} $$
Now apply the associative property of multiplication.
$$ \lim_{x \to 0} \frac{\sin x}{x} \cdot \sin x \cdot \frac{1}{1 + \cos x} $$
By the product theorem for limits, the limit of the product equals the product of the limits, provided that each limit exists. Therefore we can evaluate the three limits separately as \(x \to 0\).
$$ \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \sin x \cdot \lim_{x \to 0} \frac{1}{1 + \cos x} $$
The first limit is the standard sine limit.
$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$
The second limit tends to zero.
$$ \lim_{x \to 0} \sin x = 0 $$
The third limit tends to \( \frac{1}{2} \).
$$ \lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{2} $$
Consequently, the product of the three limits is zero, since zero is the absorbing element of multiplication. Any number multiplied by zero equals zero.
$$ \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \sin x \cdot \lim_{x \to 0} \frac{1}{1 + \cos x} = 1 \cdot 0 \cdot \frac{1}{2} = 0 $$
Therefore, the original limit also tends to zero.
$$ \lim_{x \to 0} \frac{1-\cos x}{x} = 0 $$
This completes the proof.
And so on.
