Question

$\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$

• A. 1
• B. 2
• C. 1/2
• D. 3/2

Answer 1

Answer: (B)

2

Answer 2

To evaluate the limit $\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$, we can utilize Taylor series expansions for $\tan x$ and $\sin x$ around $x=0$.

The Taylor series expansions are: $\tan x = x + \frac{x^3}{3} + O(x^5)$ $\sin x = x - \frac{x^3}{6} + O(x^5)$

First, we find the Taylor expansion for the denominator: $\tan x - \sin x = \left(x + \frac{x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{6} + O(x^5)\right)$ $\tan x - \sin x = \frac{x^3}{3} + \frac{x^3}{6} + O(x^5) = \frac{2x^3 + x^3}{6} + O(x^5) = \frac{3x^3}{6} + O(x^5) = \frac{x^3}{2} + O(x^5)$.

Next, we find the Taylor expansion for the numerator, $\tan(\tan x) - \sin(\sin x)$. For $\tan(\tan x)$: Let $u = \tan x = x + \frac{x^3}{3} + O(x^5)$. Using $\tan u = u + \frac{u^3}{3} + O(u^5)$: $\tan(\tan x) = \tan\left(x + \frac{x^3}{3} + O(x^5)\right) = \left(x + \frac{x^3}{3} + O(x^5)\right) + \frac{1}{3}\left(x + \frac{x^3}{3} + O(x^5)\right)^3 + O(x^5)$ $\tan(\tan x) = x + \frac{x^3}{3} + \frac{1}{3}(x^3) + O(x^5) = x + \frac{2x^3}{3} + O(x^5)$.

For $\sin(\sin x)$: Let $v = \sin x = x - \frac{x^3}{6} + O(x^5)$. Using $\sin v = v - \frac{v^3}{6} + O(v^5)$: $\sin(\sin x) = \sin\left(x - \frac{x^3}{6} + O(x^5)\right) = \left(x - \frac{x^3}{6} + O(x^5)\right) - \frac{1}{6}\left(x - \frac{x^3}{6} + O(x^5)\right)^3 + O(x^5)$ $\sin(\sin x) = x - \frac{x^3}{6} - \frac{1}{6}(x^3) + O(x^5) = x - \frac{2x^3}{6} + O(x^5) = x - \frac{x^3}{3} + O(x^5)$.

Now, we find the expansion for the numerator: $\tan(\tan x) - \sin(\sin x) = \left(x + \frac{2x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{3} + O(x^5)\right)$ $\tan(\tan x) - \sin(\sin x) = \frac{2x^3}{3} + \frac{x^3}{3} + O(x^5) = x^3 + O(x^5)$.

Finally, we evaluate the limit: $\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} = \lim_{x \to 0} \frac{x^3 + O(x^5)}{\frac{x^3}{2} + O(x^5)}$ Dividing the numerator and denominator by $x^3$: $\lim_{x \to 0} \frac{1 + O(x^2)}{\frac{1}{2} + O(x^2)} = \frac{1}{\frac{1}{2}} = 2$