MCQOPTIONS
Saved Bookmarks
| 1. |
Which of one of the following is correct? |
| A. | \(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\sin 4x}}{{\sin 2x}}} \right) = 1\;and\;\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\tan \;x}}{x}} \right) = 1\) |
| B. | \(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\sin 4x}}{{\sin 2x}}} \right) = \infty \;and\;\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\tan \;x}}{x}} \right) = 1\) |
| C. | \(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\sin 4x}}{{\sin 2x}}} \right) = 2\;and\;\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\tan \;x}}{x}} \right) = \infty \) |
| D. | \(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\sin 4x}}{{\sin 2x}}} \right) = 2\;and\;\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to 0} \end{array}\;\left( {\frac{{\tan \;x}}{x}} \right) = 1\) |
| Answer» E. | |