Question

The value of $\lim\limits_{x \to 0} \frac{(x-\sin 2x)(2x-\sin x)}{x^2}$ is equal to:

• A. \( 0 \)
• B. \( -2 \)
• C. \( 2 \)
• D. \( -1 \)
• E. \( -\dfrac{3}{2} \)

Hint:

For limits involving \( \sin x \) near \( x=0 \), use the expansion \( \sin x = x-\dfrac{x^3}{6}+\cdots \). It quickly reveals the leading term and makes the limit easy.

Answer 1

The Correct Option is D

Step 1: Write the given limit clearly.
We need to evaluate \[ \lim_{x\to 0}\frac{(x-\sin 2x)(2x-\sin x)}{x^2} \] As \( x \to 0 \), both brackets tend to \( 0 \), so we simplify them using standard trigonometric limits or expansions.

Step 2: Rewrite the numerator in a more convenient form.
Observe that \[ x-\sin 2x = x-2\sin x \cos x \] and \[ 2x-\sin x \] already has a simple form.
A faster way here is to use the standard small-angle approximation: \[ \sin x = x-\frac{x^3}{6}+ \cdots \] and \[ \sin 2x = 2x-\frac{(2x)^3}{6}+ \cdots = 2x-\frac{8x^3}{6}+ \cdots \]

Step 3: Simplify the first bracket \( x-\sin 2x \).
Using \[ \sin 2x = 2x-\frac{4x^3}{3}+\cdots \] we get \[ x-\sin 2x = x-\left(2x-\frac{4x^3}{3}+\cdots\right) \] \[ = -x+\frac{4x^3}{3}+\cdots \]

Step 4: Simplify the second bracket \( 2x-\sin x \).
Using \[ \sin x = x-\frac{x^3}{6}+\cdots \] we get \[ 2x-\sin x = 2x-\left(x-\frac{x^3}{6}+\cdots\right) \] \[ = x+\frac{x^3}{6}+\cdots \]

Step 5: Multiply the two simplified brackets.
Now, \[ (x-\sin 2x)(2x-\sin x) = \left(-x+\frac{4x^3}{3}+\cdots\right)\left(x+\frac{x^3}{6}+\cdots\right) \] The leading term comes from \[ (-x)(x)=-x^2 \] All remaining terms are of higher order, such as \( x^4 \) and beyond.
So, \[ (x-\sin 2x)(2x-\sin x) = -x^2 + \text{higher order terms} \]

Step 6: Divide by \( x^2 \) and take the limit.
Therefore, \[ \frac{(x-\sin 2x)(2x-\sin x)}{x^2} = \frac{-x^2+\text{higher order terms}}{x^2} \] \[ = -1+\text{terms tending to }0 \] Hence, \[ \lim_{x\to 0}\frac{(x-\sin 2x)(2x-\sin x)}{x^2}=-1 \]

Step 7: Final conclusion.
Thus, the required value of the limit is \[ \boxed{-1} \] Therefore, the correct option is \[ \boxed{(4)\ -1} \]