Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{\sin(2x) - 2\sin x}{x^3} \]

• A. \( 1 \)
• B. \( -1 \)
• C. \( 0 \)
• D. \( 2 \)

Hint:

Factor the numerator using $\sin(2x)=2\sin x\cos x$. Next use $\cos x-1=-2\sin^2(x/2)$ so the expression breaks into standard sine limits.

Answer 1

The Correct Option is B

Step 1: Use the Taylor series expansions near \( x = 0 \): \[ \sin x = x - \frac{x^3}{6} + \cdots \] \[ \sin(2x) = 2x - \frac{(2x)^3}{6} + \cdots = 2x - \frac{8x^3}{6} + \cdots \]
Step 2: Substitute into the given expression: \[ \sin(2x) - 2\sin x = \left(2x - \frac{8x^3}{6}\right) - 2\left(x - \frac{x^3}{6}\right) \]
Step 3: Simplify: \[ = 2x - \frac{8x^3}{6} - 2x + \frac{2x^3}{6} = -\frac{6x^3}{6} = -x^3 \]
Step 4: Divide by \( x^3 \) and take the limit: \[ \lim_{x \to 0} \frac{-x^3}{x^3} = -1 \]

Answer 2

Concept:

• Use exact trigonometric identities to factor the numerator instead of expanding the functions in a series.
• Then apply the standard limits $\lim_{x\to0}\dfrac{\sin x}{x}=1$ and $\lim_{x\to0}\dfrac{\sin(x/2)}{x}=\dfrac12$.

Step 1: Factor the numerator with the double-angle identity.
$\sin(2x)-2\sin x=2\sin x\cos x-2\sin x$
$=2\sin x(\cos x-1)$

Step 2: Replace $\cos x-1$ by a half-angle expression.
Since $\cos x-1=-2\sin^2(x/2)$,
$\sin(2x)-2\sin x=-4\sin x\sin^2(x/2)$

Step 3: Separate the expression into standard-limit factors.
$\dfrac{-4\sin x\sin^2(x/2)}{x^3}=-4\left(\dfrac{\sin x}{x}\right)\left(\dfrac{\sin(x/2)}{x}\right)^2$

Step 4: Apply the limits as $x\to0$.
$-4(1)\left(\dfrac12\right)^2=-4\cdot\dfrac14=-1$

Final Answer: $-1$