Question

Let $\alpha$ and $\beta$ be real numbers. If \[ \lim_{x \to 0} \frac{\tan 2x - 2 \sin \alpha x}{x(1 - \cos 2x)} = \beta, \] then $\alpha + \beta$ equals

• A. $\dfrac{1}{2}$
• B. $1$
• C. $\dfrac{3}{2}$
• D. $\dfrac{5}{2}$

Hint:

When dealing with trigonometric limits, expand all terms up to the same power of $x$ and cancel higher order terms to ensure finite limits.

Answer 1

The Correct Option is D

Step 1: Use small angle approximations. 
For $x \to 0$, \[ \tan 2x \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3}, \sin \alpha x \approx \alpha x - \frac{(\alpha x)^3}{6}. \] Also, $1 - \cos 2x \approx \frac{(2x)^2}{2} = 2x^2$. 
 

Step 2: Substitute expansions. 
\[ \frac{\tan 2x - 2\sin \alpha x}{x(1 - \cos 2x)} = \frac{2x + \frac{8x^3}{3} - 2(\alpha x - \frac{\alpha^3x^3}{6})}{x \cdot 2x^2} = \frac{2(1 - \alpha)x + \left(\frac{8}{3} + \frac{\alpha^3}{3}\right)x^3}{2x^3}. \]

Step 3: Simplify leading terms. 
For the limit to exist, the term with $x^{-2}$ must vanish, so $1 - \alpha = 0$ $\Rightarrow$ $\alpha = 1$. Substituting $\alpha = 1$, \[ \beta = \frac{\frac{8}{3} + \frac{1}{3}}{2} = \frac{9/3}{2} = \frac{3}{2}. \]

Step 4: Conclusion. 
Hence, $\alpha + \beta = 1 + \dfrac{3}{2} = \dfrac{5}{2}$.