L’Hospital’s Rule and Indeterminate Forms — Question 9

Question 9 - Solution

As $x \to 0$, we have: $$\tan x \to 0, \quad x \to 0 \quad \Rightarrow \quad \frac{\tan x - x}{x^3} \to \frac{0}{0} \quad \text{(indeterminate form)}$$

Apply L’Hospital’s Rule:

$$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2}$$

As $x \to 0$, $\sec^2 x \to 1$, so the numerator goes to 0, denominator to 0: $$\Rightarrow \frac{0}{0} \quad \text{Apply L’Hospital’s Rule again}$$

$$= \lim_{x \to 0} \frac{2 \sec^2 x \tan x}{6x} = \lim_{x \to 0} \frac{2 \cdot 1 \cdot 0}{6 \cdot 0} = \frac{0}{0} \quad \text{(Apply once more)}$$

Take third derivative:

Numerator derivative: $$\frac{d}{dx} \left[ 2 \sec^2 x \tan x \right] = 2 \left( 2 \sec^2 x \tan^2 x + \sec^4 x \right)$$

Denominator derivative: $$\frac{d}{dx} (6x) = 6$$

Now take the limit:

$$\lim_{x \to 0} \frac{2 (2 \sec^2 x \tan^2 x + \sec^4 x)}{6} = \frac{2 (2 \cdot 1 \cdot 0 + 1)}{6} = \frac{2}{6} = \boxed{\frac{1}{3}}$$