Question

plz solve part 1 of this question
..​

Answer 1

Answer:

$\large\bold\red{\frac{2 ln(2) }{\pi} }$

Step-by-step explanation:

Given,

$(i)\lim_{x \to \frac{\pi}{2} } \frac{ {2}^{ - \cos(x) } - 1}{x(x - \frac{\pi}{2}) }$

Now,

It can also be simplified as ,

$\lim_{(x - \frac{\pi}{2} )\to 0 } \frac{ {2}^{ - \cos(x) } - 1}{x(x - \frac{\pi}{2}) }$

Now,

let us consider,

$x - \frac{\pi}{2} = h \\ \\ = > x = \frac{\pi}{2} + h$

Now,

Replacing the terms,

we get,

$= \lim_{h \to 0 } \frac{ {2}^{ - \cos( \frac{\pi}{2} + h) } - 1}{( \frac{\pi}{2} + h)( \frac{\pi}{2} + h - \frac{\pi}{2} )} \\ \\ = \lim_{h \to 0 } \frac{ {2}^{ \sin(h) } - 1 }{h( \frac{\pi}{2} + h)}$

Now,

Multiplying the numerator and denominator by ( sin h ),

we get,

$= \lim_{h \to 0 } \frac{ {2}^{ \sin(h) } - 1}{ \sin(h) } \times \frac{ \sin(h) }{h( \frac{\pi}{2} + h)} \\ \\ = \lim_{h \to 0 } \frac{ {2}^{ \sin(h) } - 1}{ \sin(h) } \times\lim_{h \to 0 } \: \frac{ \sin(h) }{h( \frac{\pi}{2} + h)}$

But,

we know that,

$\lim_{x \to 0 } \frac{ {a}^{x} - 1 }{x} = ln(a)$

Therefore,

we get,

$= ln(2) \times \lim_{h \to 0 } \frac{ \sin(h) }{h( \frac{\pi}{2} + h) }$

Also,

we know that,

$\lim_{x \to 0 } \frac{ \sin(x) }{x} = 1$

Therefore,

according to this,

we get,

$= ln(2) \times \lim_{h \to 0 } \frac{1}{( \frac{\pi}{2} + h)} \\ \\ = \frac{ ln(2) }{ \frac{\pi}{2} } \\ \\ = \bold{\frac{2 ln(2) }{\pi} }$