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Question Number 190998 by mathlove last updated on 16/Apr/23

Commented by mr W last updated on 16/Apr/23

$i s i t n o r m a l w h e n s o m e b o d y a s k s$ $y o u d i r e c t l y a q u e s t i o n b u t y o u t o t a l l y$ $i r g n o r e i t w i t h o u t g i v i n g a n y r e p l y ?$ $i f i n d t h i s b e h a v i o u r a t l e a s t i m p o l i t e .$ $i m e a n f o r e x a m p l e t h e q u e s t i o n f r o m$ $T i n k u T a r a s i r i n Q 190882 . i {d o n}^{'} t$ $t h i n k y o u m u s t a n s w e r i t, b u t i t h i n k$ $y o u s h o u l d a t l e a s t g i v e a r e p l y .$ $o t h e r w i s e y o u a r e s h o w i n g t h a t y o u$ $a r e a s e l f i s h p e r s o n w h o o n l y w a n t s$ $b u t n e v e r g i v e s, w h a t i {d o n}^{'} t t h i n k i s$ $t r u e .$
Commented by mr W last updated on 16/Apr/23

$t h a n k s f o r r e p l y s i r!$ $a s i s a i d, y o u {d o n}^{'} t n e e d t o a n s w e r$ $i f y o u {d o n}^{'} t w a n t t o o r i f y o u {c a n}^{'} t$ $w r i t e i n e n g l i s h . g i v e a s h o r t r e p l y$ $l i k e ‘ ‘ s o r r y, i {c a n}^{'} t a n s w e r .^{″} i s q u i t e$ $o k . b u t j u s t s i m p l y i g n o r e i s n o t g o o d .$
Commented by mustafazaheen last updated on 16/Apr/23

$ok Mr W your speak is very accurate .$
Commented by mathlove last updated on 16/Apr/23

I don't know English, brother, I have had this problem before, but I apologize
Commented by Albert12 last updated on 16/Apr/23

$r i g h t$
Commented by mathlove last updated on 16/Apr/23

$o k$
	Answered by mustafazaheen last updated on 16/Apr/23

	$= 0$
	Commented by mathlove last updated on 16/Apr/23

	$s o l v e i t$
	Answered by mehdee42 last updated on 16/Apr/23

	$i t i s n o t i m p o s s i b l e t o {l i m}_{x \to 1^{-}} \sqrt{1 - x}$
	Answered by mr W last updated on 16/Apr/23

	$d u e t o \sqrt{x - 1}, x > 1 .$ $t h a t m e a n s \lim_{x \to 1^{-}} ◻ {d o e s n}^{'} t e x i s t$ $a n d t h e q u e s t i o n i s w r o n g .$ $a s s u m e t h e q u e s t i o n i s$ $\lim_{x \to 1^{+}} \frac{\sin (x^{3} - 1) \cos \frac{1}{1 - x}}{\sqrt{x - 1}}$ $= \lim_{x \to 1^{+}} \frac{(x - 1) (x^{2} + x + 1)}{\sqrt{x - 1}} \times \frac{\sin (x^{3} - 1)}{x^{3} - 1} \times \cos \frac{1}{1 - x}$ $= \lim_{x \to 1^{+}} \sqrt{x - 1} \times (x^{2} + x + 1) \times \frac{\sin (x^{3} - 1)}{x^{3} - 1} \times \cos \frac{1}{1 - x}$ $⩽ \lim_{x \to 1^{+}} \sqrt{x - 1} \times (x^{2} + x + 1) \times \frac{\sin (x^{3} - 1)}{x^{3} - 1} = 0 \times 1 \times 1 = 0$ $⩾ - \lim_{x \to 1^{+}} \sqrt{x - 1} \times (x^{2} + x + 1) \times \frac{\sin (x^{3} - 1)}{x^{3} - 1} = - 0 \times 1 \times 1 = 0$ $\Rightarrow \lim_{x \to 1^{+}} \frac{\sin (x^{3} - 1) \cos \frac{1}{1 - x}}{\sqrt{x - 1}} = 0$
	Commented by mehdee42 last updated on 16/Apr/23

	$t h a t s r i g h t$
	Commented by mathlove last updated on 16/Apr/23

	$t h a n k s$
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