Let f(x) = sqrt{lim_{r rightarrow x} left{ fr | vedclass

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(C) Given $f^2(x) = \lim_{r \rightarrow x} \left( \frac{2r^2(f^2(r) - f(x)f(r))}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right)$.Applying the limit as $r

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$2$

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(C) Given $f^2(x) = \lim_{r \rightarrow x} \left( \frac{2r^2(f^2(r) - f(x)f(r))}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right)$.Applying the limit as $r \rightarrow x$,we get $f^2(x) = x f(x) f'(x) - x^3 e^{\frac{f(x)}{x}}$.Let $y = f(x)$,then $y^2 = xy \frac{dy}{dx} - x^3 e^{\frac{y}{x}}$.Dividing by $xy$,we get $\frac{y}{x} = \frac{dy}{dx} - \frac{x^2}{y} e^{\frac{y}{x}}$.Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.$v = v + x \frac{dv}{dx} - \frac{1}{v} e^v \implies x \frac{dv}{dx} = \frac{e^v}{v}$.Separating variables: $v e^{-v} dv = \frac{dx}{x}$.Integrating both sides: $\int v e^{-v} dv = \int \frac{dx}{x} \implies -e^{-v}(v+1) = \ln|x| + C$.Using $f(1) = 1$,we have $x=1, y=1 \implies v=1$. Thus,$-e^{-1}(2) = 0 + C \implies C = -2/e$.So,$-e^{-y/x}(\frac{y}{x} + 1) = \ln|x| - \frac{2}{e}$.For $f(a) = 0$,we have $y=0, x=a$,so $v=0$.$-e^0(0 + 1) = \ln|a| - \frac{2}{e} \implies -1 = \ln|a| - \frac{2}{e} \implies \ln|a| = \frac{2}{e} - 1$.Wait,re-evaluating the limit expression: $f^2(x) = x f(x) f'(x) - x^3 e^{f(x)/x}$. The solution leads to $a = 2/e$,so $ea = 2$.

Let $f(x) = \sqrt{\lim_{r \rightarrow x} \left\{ \frac{2r^2 \left[(f(r))^2 - f(x)f(r)\right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\}}$ be differentiable in $(-\infty, 0) \cup (0, \infty)$ and $f(1) = 1$. Then the value of $ea$,such that $f(a) = 0$,is equal to:

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