Let f: R^{+} rightarrow R^{+} be a function s | vedclass

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(C) Given the relation $f(x f(y)) = f(x y) + x$. Since $f(x) - x = \lambda$,we have $f(x) = x + \lambda$. Substituting this into the given functional

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$\frac{2}{3}$

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(C) Given the relation $f(x f(y)) = f(x y) + x$. Since $f(x) - x = \lambda$,we have $f(x) = x + \lambda$. Substituting this into the given functional equation: $f(x(y + \lambda)) = (xy + \lambda) + x$ $x(y + \lambda) + \lambda = xy + \lambda + x$ $xy + x\lambda + \lambda = xy + \lambda + x$ Comparing the terms,we get $x\lambda = x$,which implies $\lambda = 1$. Thus,$f(x) = x + 1$. Now,we evaluate the limit: $\lim _{x \rightarrow 0} \frac{(x + 1)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1}$ Using the standard limit formula $\lim _{u \rightarrow 1} \frac{u^n - 1}{u - 1} = n$: $\lim _{x}$ ${\rightarrow 0} \frac{(1 + x)^{\frac{1}{3}} - 1}{(1 + x) - 1} \cdot \frac{(1 + x) - 1}{(1 + x)^{\frac{1}{2}} - 1} = \frac{1/3}{1/2} = \frac{2}{3}$.

Let $f: R^{+} \rightarrow R^{+}$ be a function satisfying $f(x) - x = \lambda$ (constant),$\forall x \in R^{+}$ and $f(x f(y)) = f(x y) + x, \forall x, y \in R^{+}$. Then $\lim _{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}} - 1}{(f(x))^{\frac{1}{2}} - 1} =$

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