Question

Evaluate \( \lim_{x \to 0} \frac{x - \tan(3x)}{\sin(2x)} \)

Hint:

For limits that result in \( \frac{0}{0} \), use L'Hopital's Rule by differentiating the numerator and denominator.

Answer 1

Step 1: Apply L'Hopital's Rule.
We observe that both the numerator and denominator approach 0 as \( x \to 0 \), which results in an indeterminate form \( \frac{0}{0} \). Therefore, we apply L'Hopital's Rule: \[ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \] where \( f(x) = x - \tan(3x) \) and \( g(x) = \sin(2x) \).
Step 2: Differentiate the numerator and denominator.
First, differentiate the numerator: \[ f'(x) = \frac{d}{dx} \left( x - \tan(3x) \right) = 1 - 3\sec^2(3x) \] Now, differentiate the denominator: \[ g'(x) = \frac{d}{dx} \left( \sin(2x) \right) = 2\cos(2x) \]
Step 3: Apply the limits.
Now, apply the limit as \( x \to 0 \): \[ \lim_{x \to 0} \frac{1 - 3\sec^2(3x)}{2\cos(2x)} = \frac{1 - 3\sec^2(0)}{2\cos(0)} = \frac{1 - 3(1)}{2(1)} = \frac{-2}{2} = -1 \]
Step 4: Conclusion.
The value of the limit is: \[ \boxed{-1} \]