Question

The value of $\lim_{x \to 0} \frac{x - \tan(3x)}{\sin(2x)}$ is equal to

• A. $-\frac{3}{2}$
• B. $\frac{2}{3}$
• C. -2
• D. 1
• E. -1

Hint:

Alternatively, divide terms by $x$: $\lim \frac{1 - 3\frac{\tan 3x}{3x}}{2\frac{\sin 2x}{2x}} = \frac{1 - 3(1)}{2(1)} = -1$. Standard limits are often faster than differentiation!

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to evaluate a limit that results in a \(0/0\) indeterminate form when substituting \(x = 0\).
Step 2: Key Formula or Approach:
Since it is a \(0/0\) form, we can apply L'Hôpital's Rule, which states \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\).
Alternatively, we can use standard Taylor series expansions.
Step 3: Detailed Explanation:
Let's use L'Hôpital's Rule.
Numerator: \(f(x) = x - \tan(3x)\).
Derivative: \(f'(x) = 1 - 3\sec^2(3x)\).
Denominator: \(g(x) = \sin(2x)\).
Derivative: \(g'(x) = 2\cos(2x)\).
Apply the limit to the derivatives:
\[ \lim_{x \to 0} \frac{1 - 3\sec^2(3x)}{2\cos(2x)} \] Substitute \(x = 0\):
\[ = \frac{1 - 3\sec^2(0)}{2\cos(0)} \] Since \(\sec(0) = 1\) and \(\cos(0) = 1\):
\[ = \frac{1 - 3(1)^2}{2(1)} \] \[ = \frac{1 - 3}{2} = \frac{-2}{2} = -1 \] Step 4: Final Answer:
The limit is -1.