Q71 / 180 · KCET 2017 · Mathematics
The value of \( \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} \) is
- Correct answerA\( \frac{4}{9} \)
- B\( \frac{9}{4} \)
- C\( \frac{9}{3} \)
- D\( \frac{3}{4} \)
Solution
Given that, \( \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} \)
It is a \( \frac{0}{0} \) form. Now using L' Hospital rule we get
\( \lim _{\theta \rightarrow 0} \frac{4 \sin 4 \theta}{6 \sin 6 \theta} \)
It is a \( \frac{0}{0} \) form. Now using L' Hospital rule we get \( \lim _{\theta \rightarrow 0} \frac{(4)^{2}(\cos 4 \theta)}{(6)^{2}(\cos 6 \theta)}=\frac{4 \times 4}{6 \times 6}=\frac{4}{9} \)
It is a \( \frac{0}{0} \) form. Now using L' Hospital rule we get
\( \lim _{\theta \rightarrow 0} \frac{4 \sin 4 \theta}{6 \sin 6 \theta} \)
It is a \( \frac{0}{0} \) form. Now using L' Hospital rule we get \( \lim _{\theta \rightarrow 0} \frac{(4)^{2}(\cos 4 \theta)}{(6)^{2}(\cos 6 \theta)}=\frac{4 \times 4}{6 \times 6}=\frac{4}{9} \)
Hence, the correct answer is option: 'A'.