Question:
$\lim_{x\rightarrow 2}\frac{\sin x \cos 2 - \cos x \sin 2}{x-2} =$
$\lim_{x\rightarrow 2}\frac{\sin x \cos 2 - \cos x \sin 2}{x-2} =$
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Recognizing angle addition/subtraction formulas can turn a complex-looking limit into a standard one.
Updated On: Apr 28, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Use the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$.
Step 2: Analysis
The numerator simplifies to $\sin(x-2)$. The limit becomes $\lim_{x\rightarrow2} \frac{\sin(x-2)}{x-2}$.
Step 3: Calculation
Let $u = x-2$. As $x\rightarrow2, u\rightarrow0$. $\lim_{u\rightarrow0} \frac{\sin u}{u} = 1$. Final Answer: (B)
Use the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$.
Step 2: Analysis
The numerator simplifies to $\sin(x-2)$. The limit becomes $\lim_{x\rightarrow2} \frac{\sin(x-2)}{x-2}$.
Step 3: Calculation
Let $u = x-2$. As $x\rightarrow2, u\rightarrow0$. $\lim_{u\rightarrow0} \frac{\sin u}{u} = 1$. Final Answer: (B)
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