Question

If \(\lim\limits_{x \rightarrow 0} \frac{\sin(2+x)-\sin(2-x)}{x}\)= A cos B, then the values of A and B respectively are

• A. 2, 1
• B. 2, 2
• C. 1, 1
• D. 1, 2

Hint:

When solving limits involving trigonometric functions, use standard identities like the difference of sines identity. For limits involving \(\sin(x)\) and \(x \rightarrow 0\), recall that \( \lim\limits_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \), which simplifies many trigonometric limit problems.

Answer 1

The Correct Option is B

The correct answer is (B) :2, 2.

Answer 2

The correct answer is: (B) 2, 2.

We are given the limit expression:

\( \lim\limits_{x \rightarrow 0} \frac{\sin(2+x) - \sin(2-x)}{x} \)

To solve this limit, we will use the trigonometric identity for the difference of sines:

\( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)

Substituting \( A = 2+x \) and \( B = 2-x \), we get:

\( \sin(2+x) - \sin(2-x) = 2 \cos\left(\frac{(2+x) + (2-x)}{2}\right) \sin\left(\frac{(2+x) - (2-x)}{2}\right) \)

Simplifying further:

\( = 2 \cos(2) \sin(x) \)

Now, substitute this expression back into the limit:

\( \lim\limits_{x \rightarrow 0} \frac{2 \cos(2) \sin(x)}{x} \)

Using the standard limit result \( \lim\limits_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \), we get:

\( 2 \cos(2) \)

Thus, the values of \( A \) and \( B \) are 2 and 2, respectively.

Therefore, the correct answer is (B) 2, 2.