MCQ
- A$\frac{1}{2}$
- B2
- C$\frac{1}{4}$
- D$-\frac{1}{2}$
✓
Answer
(a) $\frac{1}{2}$
Explanation: Given, $\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{x\left[\frac{\tan 2 x}{x}-1\right]}{x\left[3-\frac{\sin x}{x}\right]}$
$\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x}{2 x} \times 2-1}{3-\frac{\sin x}{x}}=\frac{1.2-1}{3-1}=\frac{2-1}{2}=\frac{1}{2}$
Explanation: Given, $\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{x\left[\frac{\tan 2 x}{x}-1\right]}{x\left[3-\frac{\sin x}{x}\right]}$
$\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x}{2 x} \times 2-1}{3-\frac{\sin x}{x}}=\frac{1.2-1}{3-1}=\frac{2-1}{2}=\frac{1}{2}$
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