Question
$$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\int\limits_2^{{{\sec }^2}\,x} {f\left( t \right)dt} }}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}}$$ equals:
A.
$$\frac{8}{\pi }f\left( 2 \right)$$
B.
$$\frac{2}{\pi }f\left( 2 \right)$$
C.
$$\frac{2}{\pi }f\left( {\frac{1}{2}} \right)$$
D.
$$4f\left( 2 \right)$$
Answer :
$$\frac{8}{\pi }f\left( 2 \right)$$
Solution :
KEY CONCEPT:
$$\eqalign{
& \frac{d}{{dx}}\left[ {\int_{g\,\left( x \right)}^{h\,\left( x \right)} {f\left( t \right)dt} } \right] \cr
& = f\left( {h\left( x \right)} \right).h'\left( x \right) - f\left( {g\left( x \right).g'\left( x \right)} \right) \cr
& = \frac{{f\left( 2 \right) \times 2 \times 2 \times 1}}{{2 \times \frac{\pi }{4}}} = \frac{8}{\pi }f\left( 2 \right) \cr
& {\text{Let }}L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\int\limits_2^{{{\sec }^2}\,x} {f\left( t \right)dt} }}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr} $$
On applying L'Hospital's rule, we get
$$\eqalign{
& L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\frac{d}{{dx}}\left[ {\int\limits_2^{{{\sec }^2}\,x} {f\left( t \right)dt} } \right]}}{{\frac{d}{{dx}}\left( {{x^2} - \frac{{{\pi ^2}}}{{16}}} \right)}}\,\,\, \cr
& L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{f\left( {{{\sec }^2}\,x} \right).2\,{{\sec }^2}\,x\,\tan \,x}}{{2x}}\,\,\, \cr} $$