Evaluate the limit as x approaches 0 of (tan x - sin x)/x³. A) 0 B) 1/2 C) 1 D) 3/2

The limit evaluates to 1/2.

Steps:

We need to evaluate the limit:

$$L = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$$

First, substitute $x=0$ into the expression:
Numerator: $\tan(0) - \sin(0) = 0 - 0 = 0$
Denominator: $0^3 = 0$
Since this is an indeterminate form $\frac{0}{0}$, we can use L'Hôpital's Rule or Taylor series expansions.

Method 1: Using L'Hôpital's Rule

Apply L'Hôpital's Rule by taking the derivative of the numerator and the denominator.

1st Application:
Derivative of numerator: $\frac{d}{dx}(\tan x - \sin x) = \sec^2 x - \cos x$
Derivative of denominator: $\frac{d}{dx}(x^3) = 3x^2$

$$L = \lim_{x \to 0} \frac{\sec^2 x - \cos x}{3x^2}$$

Substitute $x=0$:
Numerator: $\sec^2(0) - \cos(0) = 1^2 - 1 = 0$
Denominator: $3(0)^2 = 0$
Still an indeterminate form $\frac{0}{0}$.

2nd Application:
Derivative of numerator: $\frac{d}{dx}(\sec^2 x - \cos x) = 2\sec x(\sec x \tan x) - (-\sin x) = 2\sec^2 x \tan x + \sin x$
Derivative of denominator: $\frac{d}{dx}(3x^2) = 6x$

$$L = \lim_{x \to 0} \frac{2\sec^2 x \tan x + \sin x}{6x}$$

Substitute $x=0$:
Numerator: $2\sec^2(0) \tan(0) + \sin(0) = 2(1)^2(0) + 0 = 0$
Denominator: $6(0) = 0$
Still an indeterminate form $\frac{0}{0}$.

3rd Application:
Derivative of numerator: $\frac{d}{dx}(2\sec^2 x \tan x + \sin x)$
Using the product rule for $2\sec^2 x \tan x$:
$2[(2\sec x(\sec x \tan x))\tan x + \sec^2 x(\sec^2 x)] + \cos x$
$= 2[2\sec^2 x \tan^2 x + \sec^4 x] + \cos x$
$= 4\sec^2 x \tan^2 x + 2\sec^4 x + \cos x$
Derivative of denominator: $\frac{d}{dx}(6x) = 6$

$$L = \lim_{x \to 0} \frac{4\sec^2 x \tan^2 x + 2\sec^4 x + \cos x}{6}$$

Now, substitute $x=0$:
Numerator: $4\sec^2(0) \tan^2(0) + 2\sec^4(0) + \cos(0)$
$= 4(1)^2(0)^2 + 2(1)^4 + 1$
$= 0 + 2 + 1 = 3$
Denominator: 6

$$L = \frac{3}{6} = \frac{1}{2}$$

Method 2: Using Taylor Series Expansions

Recall the Maclaurin series expansions for $\sin x$ and $\tan x$ around $x=0$:
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + O(x^5)$
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots = x + \frac{x^3}{3} + O(x^5)$

Substitute these into the limit expression:

$$L = \lim_{x \to 0} \frac{\left(x + \frac{x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{6} + O(x^5)\right)}{x^3}$$ $$L = \lim_{x \to 0} \frac{x + \frac{x^3}{3} - x + \frac{x^3}{6} + O(x^5)}{x^3}$$ $$L = \lim_{x \to 0} \frac{\left(\frac{1}{3} + \frac{1}{6}\right)x^3 + O(x^5)}{x^3}$$ $$L = \lim_{x \to 0} \frac{\left(\frac{2}{6} + \frac{1}{6}\right)x^3 + O(x^5)}{x^3}$$ $$L = \lim_{x \to 0} \frac{\frac{3}{6}x^3 + O(x^5)}{x^3}$$ $$L = \lim_{x \to 0} \frac{\frac{1}{2}x^3 + O(x^5)}{x^3}$$

Divide by $x^3$:

$$L = \lim_{x \to 0} \left(\frac{1}{2} + O(x^2)\right)$$

As $x \to 0$, $O(x^2) \to 0$.

$$L = \frac{1}{2}$$

The final answer is $\boxed{\text{1/2}}$.