Let f : (0, infty) rightarrow (0, infty) be a | vedclass

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(D) Given the limit $L = \lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t - x} = 0$.Applying $L$'$H$ôpital's rule with respect to $t$:$

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$\frac{1}{e}$

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(D) Given the limit $L = \lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t - x} = 0$.Applying $L$'$H$ôpital's rule with respect to $t$:$L = \lim_{t \rightarrow x} \frac{2t f^{2}(x) - x^{2} \cdot 2f(t) f'(t)}{1} = 0$.Substituting $t = x$:$2x f^{2}(x) - 2x^{2} f(x) f'(x) = 0$.Dividing by $2x f(x)$ (since $x > 0$ and $f(x) > 0$):$f(x) - x f'(x) = 0 \Rightarrow \frac{f'(x)}{f(x)} = \frac{1}{x}$.Integrating both sides with respect to $x$:$\int \frac{f'(x)}{f(x)} dx = \int \frac{1}{x} dx \Rightarrow \ln|f(x)| = \ln|x| + C$.Since $f(x) > 0$ and $x > 0$,we have $f(x) = Cx$.Using the condition $f(1) = e$:$e = C(1) \Rightarrow C = e$.Thus,$f(x) = ex$.If $f(x) = 1$,then $ex = 1$,which implies $x = \frac{1}{e}$.

Let $f : (0, \infty) \rightarrow (0, \infty)$ be a differentiable function such that $f(1) = e$ and $\lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t - x} = 0$. If $f(x) = 1$,then $x$ is equal to

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