f(x) is continous function on R and lim_(x→1) ((f(((x+1)/x))−6)/((((x−1)/x))^2 ))=2 Evalute : lim


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Question Number 189285 by TUN last updated on 14/Mar/23

$f (x) i s c o n t i n o u s f u n c t i o n o n R$ $a n d \lim_{x \to 1} \frac{f (\frac{x + 1}{x}) - 6}{{(\frac{x - 1}{x})}^{2}} = 2$ $E v a l u t e : \lim_{x \to 1} \frac{\sqrt{f (x) + x} - x}{(x - 1)} = ¿$
	Answered by cortano12 last updated on 14/Mar/23
	$let ((x+1)/x) =u ; x=(1/(u−1)) ; (((x−1)/x))^2 =(1−(u−1))^2 L= lim_(u→2) ((f(u)−6)/((2−u)^2 )) = 2 { ((f(2)=6)),((L= lim_(u→2) )) :}(((f ′(u))/(−2(2−u)))) = 2⇒f ′(2)=0 L=lim_(u→2) (((f ′′(u))/2))=2⇒f ′′(2)=4$
	$let \frac{x + 1}{x} = u; x = \frac{1}{u - 1}; {(\frac{x - 1}{x})}^{2} = {(1 - (u - 1))}^{2}$ $L = \lim_{u \to 2} \frac{f (u) - 6}{{(2 - u)}^{2}} = 2$ ${\begin{cases} f (2) = 6 \\ L = \lim_{u \to 2} \end{cases} (\frac{f^{'} (u)}{- 2 (2 - u)}) = 2 \Rightarrow f^{'} (2) = 0$ $L = \lim_{u \to 2} (\frac{f^{″} (u)}{2}) = 2 \Rightarrow f^{″} (2) = 4$
	Answered by cortano12 last updated on 14/Mar/23

	$K = \lim_{x \to 1} \frac{\sqrt{f (x) + x} - x}{(x - 1)} =$ $[\sqrt{f (1) + 1} = 1 \Rightarrow f (1) = 0]$ $K = \lim_{x \to 1} \frac{\frac{f^{'} (x) + 1}{2 \sqrt{f (x) + x}} - 1}{1}$ $K = \frac{f^{'} (1) + 1}{2 \sqrt{f (1) + 1}} - 1 = \frac{f^{'} (1) + 1}{2} - 1$
	Commented by TUN last updated on 14/Mar/23

	$f (x) i s q u a d r a t i c f u n c t i o n$
	Commented by mr W last updated on 14/Mar/23

	$n o! t h e r e a r e i n f i n i t e p o s s i b i l i t i e s .$ $f (x) = {(x - 2)}^{2} g (x) + 6$ $g (x) c a n b e a n y f u n c t i o n w i t h g (2) = 2 .$
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