Discussion Forum

$\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan 4x}$  is equal to  

Answer:

Explanation:

 We have

  $\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan4x}$

   =  $\lim_{x \rightarrow 0}\frac{(2\sin ^{2}x(3+\cos x))}{x \times \frac{\tan 4x}{4x}\times 4x}$

   =   $\lim_{x \rightarrow 0}\frac{2\sin ^{2}x}{x^{2}}\times\lim_{x \rightarrow 0}\frac{(3+\cos x)}{4} \times\frac{1}{\lim_{x \rightarrow 0}\frac{\tan 4x}{4x}}$

 $=2\times \frac{4}{4}\times1$      $\left( \therefore \lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1 and \lim_{\theta \rightarrow 0}\frac{\tan \theta}{\theta}=1\right)$

=2