The-largest-value-of-non-negative-integer-a-for-which-lim-x-1-ax-sin-x-1-a-x-sin-x-1-1-1-x-1-x-1-4-is-

Question Number 160065 by Kunal12588 last updated on 24/Nov/21

The largest value of non - negative integer a

for which \lim_{x \to 1} {\frac{- a x + \sin (x - 1) + a}{x + \sin (x - 1) - 1}}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}

is \dots \dots . . ?

Commented by Kunal12588 last updated on 24/Nov/21

i t g i v e s a = 0, 2 b u t t h e y a r e s a y i n g a = 2 i s i n v a l i d

How is a = 2 invalid ?

Commented by Kunal12588 last updated on 24/Nov/21

h e i s a t u t o r . I t h i n k h e i s w r o n g {t h a t}^{'} s w h y

I a s k e d .

h i s r e a s o n i n g w a s L = \lim_{x \to 1} {(\frac{1 - a}{2})}^{1 + \sqrt{x}}

a n d f o r a = 2, L = \lim_{x \to 1} {(- \frac{1}{2})}^{1 + \sqrt{x}}

b u t {(n e g a t i v e)}^{\frac{o d d}{e v e n}} = n o t d e f i n e d f o r \in R

s o a s x i s a p p r o a c h i n g 1, L w i l l b e {(- \frac{1}{2})}^{\frac{1999 \dots}{100 \dots}}

w h i c h s h o u l d n o t d e f i n e d f o r r e a l n u m b e r s

Commented by Kunal12588 last updated on 24/Nov/21

so what do you think? a=2 is valid or not?

Commented by mr W last updated on 24/Nov/21

i c h a n g e d m y m i n d . t h e y a r e r i g h t .

s e e b e l o w .

Commented by Kunal12588 last updated on 24/Nov/21

s i r d o e s t h i s m e a n w h e n a = 2, L \neq \frac{1}{4}

Commented by Kunal12588 last updated on 24/Nov/21

Commented by mr W last updated on 24/Nov/21

f o r x, y \in R, w i t h a = 2

y = f (x) = {[\frac{\sin (x - 1) - 2 (x - 1)}{\sin (x - 1) + (x - 1)}]}^{1 + \sqrt{x}}

i s n o t d e f i n e d . s o l i m i t \lim_{x \to 1} f (x)

{d o e s n}^{'} t e x i s t .

i n w o l f r a m a l p h a, y \in C .

t h e r e f o r e y o u s h o u l d d e f i n e i f y o u

h a v e t o d o w i t h r e a l f u n c t i o n s o r

w i t h c o m p l e x f u n c t i o n s .

Commented by Kunal12588 last updated on 24/Nov/21

t h a n k y o u s i r

Answered by mr W last updated on 24/Nov/21

t = x - 1

L = \lim_{t \to 0} {\frac{\sin t - a t}{\sin t + t}}^{1 + \sqrt{t + 1}}

L = {(\frac{1 - a}{2})}^{2}

\frac{1 - a}{2} ⩾ 0 \Rightarrow a ⩽ 1 *)

{(\frac{1 - a}{2})}^{2} = \frac{1}{4}

\Rightarrow a = 0, 2 (r e j e c t e d d u e t o *))

*)

f {(t)}^{g (t)} i s d e f i n e d o n l y f o r f (t) ⩾ 0

i f g (t) \neq e v e n

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