Correct Answer - Option 1 : 1/4
Concpet:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)
II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule as:
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Calculation:
\(\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left[ {\frac{{1 - \sin 2x}}{{1 + \cos 4x}}} \right] = \frac{0}{0}\)
Since this is an indeterminate form, we can apply the L-Hospital Rule as:
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
\(\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left[ {\frac{{1 - \sin 2x}}{{1 + \cos 4x}}} \right] = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left[ {\frac{{ - 2\cos 2x}}{{ - \;4\sin 4x}}} \right] = \frac{0}{0}\)
Since this is also 0/0 form, we again apply the L-Hospital Rule as:
\(\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left[ {\frac{{ - 2\cos 2x}}{{ - \;4\sin 4x}}} \right] = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left[ {\frac{{4\sin 2x}}{{ - \;16\cos 4x}}} \right] = \frac{1}{4}\)
Hence, option A is the correct answer.
