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:
When the limit results in an indeterminate form like $0/0$, we can use standard limits ($\lim_{x \to 0} \frac{\sin ax}{x} = a$) or apply L'Hopital's Rule.

Step 2: Detailed Explanation:
As $x \to 0$, the expression $\frac{0 - \tan(0)}{\sin(0)} = \frac{0}{0}$.
Applying L'Hopital's Rule (differentiating numerator and denominator):
\[ \lim_{x \to 0} \frac{\frac{d}{dx}(x - \tan(3x))}{\frac{d}{dx}(\sin(2x))} \]
\[ = \lim_{x \to 0} \frac{1 - 3\sec^2(3x)}{2\cos(2x)} \]
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 3: Final Answer:
The limit is -1.