Question

Please answer me fast ​

Answer 1

Step-by-step explanation:

Since, the given function is continuous at x = 0, so LHL and RHL at x=0 will be equal,

So,

$LHL = \lim_{h \rarr0} f(0 - h) = \lim_{h \rarr0}a \sin \{ \frac{\pi}{2} ( - h + 1) \}$

$= \lim _{h \rarr0}a \sin( \frac{\pi}{2} ) = a$

$RHL = \lim_{ h \rarr0} f(0 + h) = \lim_{h \rarr0} \frac{ \tan(h) - \sin(h) }{ {h}^{3} }$

$= \lim_{h \rarr0} \frac{ \sec ^{2} (h) - \cos(h) }{ 3{h}^{2} }$

$= \lim_{h \rarr0} \frac{2 \sec ^{2} (h) \tan(h) + \sin(h) }{ 6h }$

$= \lim_{h \rarr0} \frac{2 \sec ^{2} (h) \tan(h) }{ 6h } + \lim _{h \rarr0} \frac{ \sin(h) }{6h}$

$= \frac{1}{3} \lim_{h \rarr0} \frac{\tan(h) }{ h } + \frac{1}{6} \lim _{h \rarr0} \frac{ \sin(h) }{h}$

$= \frac{1}{2}$

Since, LHL = RHL, so

$a= \frac{1}{2}$