MCQ
Evaluate the following limit : $ \displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
- A$1$
- B$3$
- ✓$9$
- D$0$
✓
Answer
Correct option: C.
$9$
$\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
$=\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{x}} \times\frac{\sin 3\text{x}}{\text{x}}$
$= \displaystyle\lim_{\text{x} \rightarrow 0} 3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg) \times3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg)$
$=3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}\times3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}$
$ = 3\times3$
$=9$
$=\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{x}} \times\frac{\sin 3\text{x}}{\text{x}}$
$= \displaystyle\lim_{\text{x} \rightarrow 0} 3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg) \times3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg)$
$=3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}\times3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}$
$ = 3\times3$
$=9$
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