L’Hospital’s Rule and Indeterminate Forms — Question 7

Question 7 - Solution

As $x \to 0$, we know: $$\tan x \to 0, \quad x \to 0 \Rightarrow \tan x - x \to 0 \quad \text{and} \quad x^3 \to 0 \Rightarrow \text{Indeterminate form } \frac{0}{0}$$

Apply L’Hospital’s Rule:

First derivative: $$\lim_{x \to 0} \frac{\frac{d}{dx}[\tan x - x]}{\frac{d}{dx}[x^3]} = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2}$$

As $x \to 0$, $\sec^2 x \to 1$, so numerator $\to 0$, denominator $\to 0$, still $\frac{0}{0}$

Second derivative: $$\lim_{x \to 0} \frac{\frac{d}{dx}[\sec^2 x - 1]}{\frac{d}{dx}[3x^2]} = \lim_{x \to 0} \frac{2\sec^2 x \cdot \sec x \tan x}{6x}$$

As $x \to 0$, $\tan x \to 0$, so numerator $\to 0$, denominator $\to 0$

Third derivative: $$\text{Differentiate numerator and denominator again. Let’s simplify instead:}$$

Use the series: $$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots \Rightarrow \tan x - x = \frac{x^3}{3} + \cdots$$

Then: $$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} + \cdots}{x^3} = \frac{1}{3}$$

Final Answer: $\boxed{\dfrac{1}{3}}$