Question:

Step 1: Simplify the first term of the expression. The first term is given by: $$ \frac{(\tan x - \sin x)}{(\tan x - \sin x) + (\tan x + \sin x)} $$ Simplify the denominator: $$ (\tan x - \sin x) + (\tan x + \sin x) = 2 \tan x $$ So the first term becomes: $$ \frac{\tan x - \sin x}{2 \tan x} $$ Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$: $$ \frac{\frac{\sin x}{\cos x} - \sin x}{2 \frac{\sin x}{\cos x}} $$ Factor out $\sin x$ from the numerator: $$ \frac{\sin x \left(\frac{1}{\cos x} - 1\right)}{2 \frac{\sin x}{\cos x}} $$ For $x \to 0$, $\sin x \neq 0$ for $x$ in a neighborhood of 0, allowing cancellation of $\sin x$: $$ \frac{\frac{1}{\cos x} - 1}{\frac{2}{\cos x}} $$ Multiply the numerator and denominator by $\cos x$: $$ \frac{1 - \cos x}{2} $$ Step 2: Evaluate the limit of the first term as $x \to 0$. $$ \lim_{x \to 0} \frac{1 - \cos x}{2} $$ Substitute $x=0$: $$ = \frac{1 - \cos 0}{2} = \frac{1 - 1}{2} = \frac{0}{2} = 0 $$ Step 3: Determine the value of the entire expression. The given expression is a sum of two terms: $$ \lim_{x \to 0} \left( \frac{(\tan x - \sin x)}{(\tan x - \sin x) + (\tan x + \sin x)} + \frac{1}{r^3 - r} \right) $$ The limit of a sum is the sum of the limits, provided they exist: $$ = \lim_{x \to 0} \frac{(\tan x - \sin x)}{(\tan x - \sin x) + (\tan x + \sin x)} + \lim_{x \to 0} \frac{1}{r^3 - r} $$ From Step 2, the limit of the first term is $0$. The second term, $\frac{1}{r^3 - r}$, is a constant with respect to $x$. Therefore, its limit as $x \to 0$ is simply its value. Thus, the total limit is: $$ 0 + \frac{1}{r^3 - r} = \frac{1}{r^3 - r} $$ The value of the expression is $3$.