Question

$\large \displaystyle \rm \red{ \lim_{h \to0} \frac{1}{h} \int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{4} + h } \frac{ \sin(x) }{x} \: dx}$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \rm {\red{ \lim_{h \to0} \dfrac{1}{h} \int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{4} + h } \frac{ \sin(x) }{x} \: dx}}$

$\displaystyle\rm { = \lim_{h \to0} \dfrac{ \displaystyle\int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{4} + h } \frac{ \sin(x) }{x} \: dx}{h} }$

$\displaystyle\rm { = \lim_{h \to0} \dfrac{ \displaystyle \rm\dfrac{d}{dh}\int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{4} + h } \frac{ \sin(x) }{x} \: dx}{ \dfrac{d}{dh}(h)} }$

$\displaystyle\rm { = \lim_{h \to0} \dfrac{ \displaystyle \rm\dfrac{ \sin \left( \dfrac{\pi}{4} + h\right) }{\dfrac{\pi}{4} + h} \cdot\dfrac{d}{dh}\left( \dfrac{\pi}{4} + h\right) + 0}{ \dfrac{d}{dh}(h)} }$

$\displaystyle\rm { = \lim_{h \to0} \dfrac{ \displaystyle \rm\dfrac{ \sin \left( \dfrac{\pi}{4} + h\right) }{\dfrac{\pi}{4} + h} \cdot1}{ 1} }$

$\displaystyle\rm { = \lim_{h \to0} \dfrac{ \sin \left( \dfrac{\pi}{4} + h\right) }{\dfrac{\pi}{4} + h} }$

$\rm { =\dfrac{ \sin \left( \dfrac{\pi}{4}\right) }{\dfrac{\pi}{4} } }$

$\rm { =\dfrac{ 4 \cdot\dfrac{1}{ \sqrt{2} } }{\pi } }$

$\rm { =\dfrac{ 2\sqrt{2} }{\pi } }$