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Question Number 34375 by rahul 19 last updated on 05/May/18

T h e v a l u e o f

{l i m}_{x \to \frac{π}{2}} \frac{[\frac{x}{2}]}{\log (\sin x)} = ?

[.] = g r e a t e s t i n t e g e r f u n c t i o n .

Answered by MJS last updated on 05/May/18

f (x) = [\frac{x}{2}] = 0 if x is close to \frac{π}{2}

so f^{'} (x) = 0

g (x) = \ln \sin x

g^{'} (x) = \frac{1}{\tan x}

\frac{f^{'} (x)}{g^{'} (x)} = 0 \times \tan x = 0

\lim_{x \to \frac{π}{2}} \frac{[\frac{x}{2}]}{\ln \sin x} = \underset{x \to \frac{π}{2}}{\lim 0} \times \tan x = 0

Commented by rahul 19 last updated on 05/May/18

T h a n k y o u s i r .