Limit of 1 − cos x as x approaches 0
Evaluate the limit of $f(x) = 1-\cos(x)$ as $x \to 0$.
Explanation of Method
Since $\cos$ is continuous everywhere, the limit of $1-\cos(x)$ as $x\to0$ is found immediately by direct substitution — no algebraic manipulation is strictly necessary. What is more interesting, and often useful in larger problems, is exactly how fast $f(x)$ approaches $0$; the Taylor expansion of $\cos x$ reveals that $1-\cos x$ behaves like $\frac{x^2}{2}$ for small $x$, a fact frequently used as a building block in more complex limits.
- Evaluate the limit directly using continuity of $\cos$.
- Expand $\cos x$ with its Taylor series to see the precise rate at which $f(x)\to0$.
Step 1 · Direct evaluation using continuity
Since $\cos$ is continuous at every real number, in particular at $0$:
$$ \lim_{x\to0}\cos(x) = \cos(0) = 1. $$
Therefore:
$$ \lim_{x\to0} f(x) = \lim_{x\to0}\left(1-\cos x\right) = 1-1 = 0. $$
Step 2 · The rate of vanishing (Taylor expansion)
The Taylor series for $\cos x$ about $x=0$ is:
$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots. $$
Substituting into $f(x)$:
$$ f(x) = 1-\cos x = 1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\right) = \frac{x^2}{2} - \frac{x^4}{24} + \cdots. $$
As $x\to0$, the higher-order terms ($x^4$ and beyond) vanish faster than $x^2$, so:
$$ f(x) \sim \frac{x^2}{2} \qquad \text{for } x \text{ near } 0, $$
confirming $\lim_{x\to0}f(x)=\lim_{x\to0}\frac{x^2}{2}=0$, consistent with Step 1.
Summary Table
| Approach | Key fact used | Result |
|---|---|---|
| Direct substitution | Continuity of $\cos$ at $0$ | $0$ |
| Taylor expansion | $1-\cos x \sim \dfrac{x^2}{2}$ | $0$, with quadratic rate |
Additional Notes
- Because $f(0)=1-\cos(0)=0$ matches the limit, $f$ is not just "approaching" $0$ — it is exactly continuous at $x=0$ (indeed, everywhere), so this limit is really testing recognition of that fact rather than requiring any indeterminate-form technique.
- The asymptotic relation $1-\cos x \sim \frac{x^2}{2}$ is one of the most frequently used small-angle approximations in calculus, showing up repeatedly when simplifying more complicated limits that contain a factor of $1-\cos(\cdot)$.
- Geometrically, $1-\cos x$ measures how far the point $(\cos x,\sin x)$ on the unit circle has dropped below the topmost point $(1,0)$'s horizontal level as the angle $x$ moves away from $0$ — this drop is quadratically small for small angles, which is exactly what the Taylor expansion captures.
Result
The value of the limit is: