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Question Number 189285 by TUN last updated on 14/Mar/23
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)))=¿
$${f}\left({x}\right)\:{is}\:{continous}\:{function}\:{on}\:{R} \\ $$$${and}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{f}\left(\frac{{x}+\mathrm{1}}{{x}}\right)−\mathrm{6}}{\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$${Evalute}\::\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{{f}\left({x}\right)+{x}}−{x}}{\left({x}−\mathrm{1}\right)}=¿ \\ $$
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
$$\:\mathrm{let}\:\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}}\:=\mathrm{u}\:;\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{u}−\mathrm{1}}\:;\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} =\left(\mathrm{1}−\left(\mathrm{u}−\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{f}\left(\mathrm{u}\right)−\mathrm{6}}{\left(\mathrm{2}−\mathrm{u}\right)^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\:\begin{cases}{\mathrm{f}\left(\mathrm{2}\right)=\mathrm{6}}\\{\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\mathrm{2}} {\mathrm{lim}}}\end{cases}\left(\frac{\mathrm{f}\:'\left(\mathrm{u}\right)}{−\mathrm{2}\left(\mathrm{2}−\mathrm{u}\right)}\right)\:=\:\mathrm{2}\Rightarrow\mathrm{f}\:'\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\mathrm{L}=\underset{\mathrm{u}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\frac{\mathrm{f}\:''\left(\mathrm{u}\right)}{\mathrm{2}}\right)=\mathrm{2}\Rightarrow\mathrm{f}\:''\left(\mathrm{2}\right)=\mathrm{4} \\ $$
Answered by cortano12 last updated on 14/Mar/23
K=lim_(x→1) (((√(f(x)+x))−x)/((x−1))) = [ (√(f(1)+1))=1⇒f(1)=0 ] K= lim_(x→1) ((((f ′(x)+1)/(2(√(f(x)+x)))) −1)/1) K= ((f ′(1)+1)/(2(√(f(1)+1))))−1=((f ′(1)+1)/2)−1
$$\mathrm{K}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{x}}−\mathrm{x}}{\left(\mathrm{x}−\mathrm{1}\right)}\:= \\ $$$$\:\left[\:\sqrt{\mathrm{f}\left(\mathrm{1}\right)+\mathrm{1}}=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{0}\:\right]\: \\ $$$$\mathrm{K}=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\frac{\mathrm{f}\:'\left(\mathrm{x}\right)+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{x}}}\:−\mathrm{1}}{\mathrm{1}} \\ $$$$\mathrm{K}=\:\frac{\mathrm{f}\:'\left(\mathrm{1}\right)+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{f}\left(\mathrm{1}\right)+\mathrm{1}}}−\mathrm{1}=\frac{\mathrm{f}\:'\left(\mathrm{1}\right)+\mathrm{1}}{\mathrm{2}}−\mathrm{1} \\ $$
Commented by TUN last updated on 14/Mar/23
f(x) is quadratic function
$${f}\left({x}\right)\:{is}\:{quadratic}\:{function} \\ $$
Commented by mr W last updated on 14/Mar/23
no! there are infinite possibilities. f(x)=(x−2)^2 g(x)+6 g(x) can be any function with g(2)=2.
$${no}!\:{there}\:{are}\:{infinite}\:{possibilities}. \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{2}} {g}\left({x}\right)+\mathrm{6}\: \\ $$$${g}\left({x}\right)\:{can}\:{be}\:{any}\:{function}\:{with}\:{g}\left(\mathrm{2}\right)=\mathrm{2}. \\ $$

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