$\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\} =$

  • A
    $\frac{1}{16}$
  • B
    $\frac{1}{32}$
  • C
    $\frac{1}{64}$
  • D
    $\frac{1}{8}$

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Similar Questions

$\lim _{x \rightarrow 0} \frac{x \tan 4x - 2x \tan 2x}{(1 - \cos 4x)^2} = $

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ ની કિંમત $.......$ છે.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + \sin 6x}}{{\sin 5x - \sin 3x}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}} = $

જો $n < m$ આપેલ હોય,તો $\lim _{x \rightarrow 0} \frac{\sin (x^m)}{(\sin x)^n}$ ની કિંમત શોધો.

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