Find the limit of (tan x - sin x) / x^2 as x approaches 0.

Step-by-step solution

1. 1

   Do not replace $\tan x$ and $\sin x$ by $x$ separately. Both are equivalent to $x$ near $0$, so the shortcut suggests a numerator of $x-x=0$ and an answer of $0$ "for the wrong reason". Equivalent infinitesimals may be substituted in products and quotients, never in a difference where the leading terms cancel. The real size of the numerator must be found first.

2. 2

   Factor $\sin x$ out of the numerator. Writing $\tan x=\dfrac{\sin x}{\cos x}$:

   $\tan x-\sin x=\sin x\left(\frac{1}{\cos x}-1\right)=\sin x\cdot\frac{1-\cos x}{\cos x}$

   This exact rewrite replaces a dangerous difference with a product of three well-understood pieces.

3. 3

   Split the quotient into standard limits. Dividing by $x^2=x\cdot x$ and pairing each factor with one $x$:

   $\frac{\tan x-\sin x}{x^2}=\frac{\sin x}{x}\cdot\frac{1-\cos x}{x}\cdot\frac{1}{\cos x}$

   Now each factor is a textbook limit, and the product rule for limits applies because all three converge.

4. 4

   Evaluate the three factors.

   $\frac{\sin x}{x}\to 1,\qquad \frac{1-\cos x}{x}\to 0,\qquad \frac{1}{\cos x}\to 1$

   The middle one is the decisive factor: $1-\cos x\approx\tfrac{x^2}{2}$, so dividing by a single $x$ still leaves something that vanishes.

5. 5

   Multiply the limits.

   $1\cdot 0\cdot 1=0$

6. 6

   Cross-check with Taylor series. From $\tan x=x+\tfrac{x^3}{3}+O(x^5)$ and $\sin x=x-\tfrac{x^3}{6}+O(x^5)$, the difference is $\tfrac{x^3}{3}+\tfrac{x^3}{6}=\tfrac{x^3}{2}$. Dividing by $x^2$ leaves $\tfrac{x}{2}\to 0$. Numerically the quotient is $0.0501$ at $x=0.1$ and $0.0050$ at $x=0.01$ — shrinking by a factor of ten each time, exactly like $x/2$.