Question

Evaluate: \[ \lim_{x \to 0} \frac{\sin 3x - 3 \sin x}{x^3} \]

• A. $0$
• B. $-4$
• C. $-9$
• D. $4.5$

Hint:

Use series expansion or repeated L'Hospital's rule to evaluate limits involving indeterminate forms.

Answer 1

The Correct Option is B

To solve the problem, we need to evaluate the limit:

$$ \lim_{x \to 0} \frac{\sin 3x - 3\sin x}{x^3} $$

1. Understanding the Approach:
We are given a limit expression involving $\sin 3x$ and $\sin x$. As $x \to 0$, both $\sin 3x$ and $3\sin x$ approach $0$, so the numerator tends to $0$, and the denominator also tends to $0$. This is an indeterminate form of the type $\frac{0}{0}$, which suggests we need to use either series expansion or repeated differentiation (L'Hôpital's Rule). Here, we'll use the Maclaurin series expansion to simplify.

2. Maclaurin Series Expansion:
Let us expand both $\sin 3x$ and $\sin x$ using their Taylor series around $x = 0$: 

$\sin 3x = 3x - \frac{(3x)^3}{6} + \frac{(3x)^5}{120} - \cdots = 3x - \frac{27x^3}{6} + \frac{243x^5}{120} - \cdots$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$
So, 
$3\sin x = 3x - \frac{3x^3}{6} + \frac{3x^5}{120} - \cdots = 3x - \frac{3x^3}{6} + \frac{3x^5}{120}$

3. Subtract the Series:
Now compute the numerator: 

$\sin 3x - 3\sin x = (3x - \frac{27x^3}{6} + \frac{243x^5}{120}) - (3x - \frac{3x^3}{6} + \frac{3x^5}{120})$
$= 3x - 3x - \left(\frac{27x^3}{6} - \frac{3x^3}{6}\right) + \left(\frac{243x^5}{120} - \frac{3x^5}{120}\right)$
$= -\frac{24x^3}{6} + \frac{240x^5}{120}$
$= -4x^3 + 2x^5$

4. Divide by $x^3$:
Now plug back into the limit: 

$\lim_{x \to 0} \frac{-4x^3 + 2x^5}{x^3} = \lim_{x \to 0} (-4 + 2x^2)$

5. Final Evaluation:
As $x \to 0$, $2x^2 \to 0$, so the limit becomes: 

$\lim_{x \to 0} (-4 + 0) = -4$

Final Answer:
The value of the limit is $ {-4} $.