MCQ
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin7\text{x}}{\sin3\text{x}}$ equals:
- A$\frac{7}{3}$
- B$\frac{10}{3}$
- C$\frac{14}{3}$
- D$\frac{1}{3}$
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The area of a triangle with vertices at (-4, -1), (1, 2) and (4, -3) is:
If $\text{f}(\text{x})=\text{x}-[\text{x}],\in\text{R}$ then $\text{f}'\big(\frac{1}{2}\big)$ is equal to:
$\frac{3}{2}$
$1$
$0$
$-1$