f-x-is-continous-function-on-R-and-lim-x-1-f-x-1-x-6-x-1-x-2-2-Evalute-lim-x-1-f-x-x-x-x-1-

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 \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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