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Step-by-Step Solution
Step 1: Understand the Given Conditions
We have a function $f:(0,\infty)\to(0,\infty)$ such that:
$f(1) = e$
$\displaystyle \lim_{t \to x} \frac{t^2 f^2(x)\;-\;x^2 f^2(t)}{t - x} \;=\; 0$ for all $x > 0$.
We need to find the value of $x$ for which $f(x) = 1$.
Step 2: Rewrite the Limit in a More Usable Form
The expression
$$
\lim_{t \to x} \frac{t^2 f^2(x) \;-\; x^2 f^2(t)}{t - x}
$$
resembles the difference quotient for a derivative, suggesting we might try L'Hospital's Rule or an equivalent derivative-based argument.
Step 3: Apply L'Hospital's Rule
By treating $t$ as the variable tending to $x$, we differentiate the numerator and denominator with respect to $t$. The denominator becomes 1 upon differentiation. For the numerator:
Differentiate $t^2 f^2(x)$ with respect to $t$:
$$
\frac{d}{dt}\bigl(t^2 f^2(x)\bigr) = 2\,t \,f^2(x)
$$
since $f^2(x)$ is constant with respect to $t$.
Differentiate $-\,x^2 f^2(t)$ with respect to $t$:
$$
\frac{d}{dt}\bigl(-\,x^2 f^2(t)\bigr) = -\,x^2\,\frac{d}{dt}\bigl(f^2(t)\bigr) = -\,x^2 \cdot 2 f(t) f'(t).
$$
Hence, applying L'Hospital's Rule to the given limit yields:
$$
\lim_{t \to x} \frac{t^2 f^2(x)\;-\;x^2 f^2(t)}{t - x}
\;=\;
\lim_{t \to x} \Bigl( 2\,t \,f^2(x) \;-\; 2\,x^2\,f(t)\,f'(t) \Bigr).
$$
Since the limit is given as 0, we evaluate the resulting expression at $t = x$:
$$
2\,x\,f^2(x) \;-\; 2\,x^2\,f(x)\,f'(x) = 0.
$$
Step 4: Simplify the Expression
Factor out $2x\,f(x)$:
$$
2x\,f^2(x) \;-\; 2x^2\,f(x)\,f'(x)
=
2x\,f(x)\Bigl(f(x) \;-\; x\,f'(x)\Bigr)
=
0.
$$
Since $x > 0$ and $f(x) > 0$ (given by the domain and range of $f$), we conclude:
$$
f(x) \;-\; x\,f'(x) = 0
\quad \Longrightarrow \quad
f(x) = x\,f'(x).
$$
Step 5: Identify the Differential Equation
From $f(x) = x\,f'(x)$, we can rewrite it as:
$$
\frac{f'(x)}{f(x)} = \frac{1}{x}.
$$
This is a simple separable differential equation.
Step 6: Solve the Differential Equation
Separate the variables and integrate:
$$
\int \frac{1}{f(x)}\,df(x) \;=\; \int \frac{1}{x}\,dx.
$$
This gives:
$$
\ln\bigl(f(x)\bigr) \;=\; \ln(x) \;+\; \ln(C),
$$
where $C$ is the constant of integration. Hence,
$$
f(x) = C\,x.
$$
Step 7: Use the Initial Condition $f(1) = e$
We know $f(1) = e$, so substitute $x=1$ into $f(x) = Cx$:
$$
f(1) = C \cdot 1 = C \quad\Longrightarrow\quad C = e.
$$
Therefore,
$$
f(x) = e\,x.
$$
Step 8: Find $x$ Such That $f(x) = 1$
We need to solve $f(x) = 1$:
$$
e\,x = 1
\;\;\Longrightarrow\;\;
x = \frac{1}{e}.
$$
Thus, the value of $x$ for which $f(x)=1$ is
$$
\boxed{\frac{1}{e}}.
$$
