Limit of tan x − sin x as x approaches 0
Evaluate the limit of $\tan x - \sin x$ as $x \to 0$.
Explanation of Method
Since $\tan$ and $\sin$ are both continuous at $x=0$ and both equal $0$ there, their difference tends to $0$ immediately by continuity. As with the earlier problems involving $1-\cos x$ and $\sinh x-x$, the more informative question is the precise rate at which this difference vanishes; the Taylor expansions of $\tan x$ and $\sin x$ share the same linear term, so their difference is governed entirely by the next surviving power, $x^3$.
- Evaluate the limit directly using continuity of $\tan$ and $\sin$.
- Expand $\tan x$ and $\sin x$ with their Taylor series to find the precise rate at which the difference vanishes.
Step 1 · Direct evaluation using continuity
Since $\tan$ and $\sin$ are both continuous at $x=0$, with $\tan(0)=0$ and $\sin(0)=0$:
$$ \lim_{x\to0}\left(\tan x - \sin x\right) = 0 - 0 = 0. $$
Step 2 · The rate of vanishing (Taylor expansion)
The Taylor series for $\tan x$ and $\sin x$ about $x=0$ are:
$$ \tan x = x+\frac{x^3}{3}+O(x^5), \qquad \sin x = x-\frac{x^3}{6}+O(x^5). $$
Subtracting:
$$ \tan x - \sin x = \left(x+\frac{x^3}3\right) - \left(x-\frac{x^3}6\right) + O(x^5) = \left(\frac{x^3}3+\frac{x^3}6\right)+O(x^5). $$
Combining the fractions over a common denominator of $6$:
$$ \frac{x^3}{3}+\frac{x^3}{6} = \frac{2x^3}{6}+\frac{x^3}{6} = \frac{3x^3}{6} = \frac{x^3}{2}. $$
So:
$$ \tan x - \sin x \sim \frac{x^3}{2} \qquad \text{for } x \text{ near } 0, $$
confirming $\lim_{x\to0}(\tan x-\sin x)=\lim_{x\to0}\frac{x^3}2=0$, consistent with Step 1.
Summary Table
| Approach | Key fact used | Result |
|---|---|---|
| Direct substitution | Continuity of $\tan$ and $\sin$ at $0$ | $0$ |
| Taylor expansion | $\tan x-\sin x \sim \dfrac{x^3}{2}$ | $0$, with cubic rate |
Additional Notes
- The linear terms of $\tan x$ and $\sin x$ (both equal to $x$) cancel exactly, which is why the difference vanishes at all — the surviving behavior comes entirely from the next terms in each series, $\frac{x^3}3$ and $-\frac{x^3}6$, whose difference gives the coefficient $\frac12$.
- This cubic rate, $\tan x-\sin x\sim\frac{x^3}2$, is a standard building block for evaluating more complex limits that contain $\tan x-\sin x$ as a factor — for instance, dividing by $x^3$ in a larger expression would immediately reveal a factor of $\frac12$.
- Geometrically, $\tan x-\sin x = \sin x\left(\frac1{\cos x}-1\right) = \sin x\cdot\frac{1-\cos x}{\cos x}$, combining the known rates $\sin x\sim x$ and $1-\cos x\sim\frac{x^2}2$ (with $\cos x\to1$ in the denominator) to reproduce the same $\frac{x^3}2$ rate independently.
Result
The value of the limit is: