MCQ
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$ is equal to:
  • $\frac{4}{9}$
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • D
    $-1$

Answer

Correct option: A.
$\frac{4}{9}$
Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$
$=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$
$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]$
$=\Big[\frac{2\theta}{2\theta}\Big]^{2}$
$=\Big(\frac{2}{3}\Big)^{2}$
$=\frac{4}{9}$

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