If f(x)=( x)-( x) x- x is continuous at x=0, then f(0) is equal to:

Question

If $f(x)=\dfrac{	an(	an x)-\sin(\sin x)}{	an x-\sin x}$ is continuous at $x=0$, then $f(0)$ is equal to:

Accepted Answer

Split the numerator: $\tan(\tan x)-\sin(\sin x)=[\tan(\tan x)-\tan x]+[\tan x-\sin x]+[\sin x-\sin(\sin x)]$.

Using the standard small-$u$ approximations $\tan u-u\approx\dfrac{u^3}{3}$ and $u-\sin u\approx\dfrac{u^3}{6}$:

- $\tan(\tan x)-\tan x\approx\dfrac{(\tan x)^3}{3}\approx\dfrac{x^3}{3}$
- $\tan x-\sin x=\tan x(1-\cos x)\approx x\cdot\dfrac{x^2}{2}=\dfrac{x^3}{2}$
- $\sin x-\sin(\sin x)\approx\dfrac{(\sin x)^3}{6}\approx\dfrac{x^3}{6}$

Adding: numerator $\approx x^3\left(\dfrac13+\dfrac12+\dfrac16\right)=x^3(1)=x^3$, while the denominator $\tan x-\sin x\approx\dfrac{x^3}{2}$.

$$f(0)=\lim_{x\to0}\frac{x^3}{x^3/2}=2$$

Solutions · 1