Key Concepts and Formulas

• Definition of the Derivative: The derivative of a function $f(x)$ at a point $x$ is defined as $f'(x) = \mathop {\lim }\limits_{t \to x} \frac{f(t) - f(x)}{t - x}$.
• L'Hôpital's Rule: If a limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ is encountered, L'Hôpital's Rule can be applied. It states that $\mathop {\lim }\limits_{t \to x} \frac{g(t)}{h(t)} = \mathop {\lim }\limits_{t \to x} \frac{g'(t)}{h'(t)}$, provided the latter limit exists.
• Separation of Variables: A first-order differential equation of the form $\frac{dy}{dx} = g(x)h(y)$ can be solved by separating variables: $\frac{dy}{h(y)} = g(x)dx$.
• Integration of $\frac{1}{x}$: The integral of $\frac{1}{x}$ with respect to $x$ is $\ln|x| + C$.

Step-by-Step Solution

Step 1: Analyze the given limit expression. We are given the limit: $\mathop {\lim }\limits_{t \to x} \frac{t^2 f^2(x) - x^2 f^2(t)}{t - x} = 0$ As $t \to x$, the numerator approaches $x^2 f^2(x) - x^2 f^2(x) = 0$, and the denominator approaches $x - x = 0$. This is an indeterminate form of type $\frac{0}{0}$.

Step 2: Apply L'Hôpital's Rule to evaluate the limit. We differentiate the numerator and the denominator with respect to $t$ (treating $x$ as a constant for differentiation purposes with respect to $t$). The derivative of the numerator $N(t) = t^2 f^2(x) - x^2 f^2(t)$ with respect to $t$ is: $N'(t) = \frac{d}{dt}(t^2 f^2(x)) - \frac{d}{dt}(x^2 f^2(t))$ $N'(t) = 2t f^2(x) - x^2 (2 f(t) f'(t))$ $N'(t) = 2t f^2(x) - 2x^2 f(t) f'(t)$

The derivative of the denominator $D(t) = t - x$ with respect to $t$ is: $D'(t) = \frac{d}{dt}(t - x) = 1$

Applying L'Hôpital's Rule, the limit becomes: $\mathop {\lim }\limits_{t \to x} \frac{2t f^2(x) - 2x^2 f(t) f'(t)}{1} = 0$

Step 3: Substitute $t=x$ into the resulting expression. Since the limit exists and is equal to 0, we can substitute $t=x$ into the expression after applying L'Hôpital's Rule: $2x f^2(x) - 2x^2 f(x) f'(x) = 0$

Step 4: Simplify the equation and derive the differential equation. We can factor out $2x f(x)$ from the equation: $2x f(x) [f(x) - x f'(x)] = 0$ We are given that the domain and codomain of $f$ are $(0, \infty)$. This means $x > 0$ and $f(x) > 0$. Therefore, $2x f(x) \neq 0$. For the entire expression to be zero, the term in the bracket must be zero: $f(x) - x f'(x) = 0$ Rearranging this equation, we get: $f(x) = x f'(x)$

Step 5: Solve the differential equation. We can rewrite the equation as: $f'(x) = \frac{f(x)}{x}$ This is a first-order linear differential equation. We can separate the variables: $\frac{f'(x)}{f(x)} = \frac{1}{x}$ Or, in differential notation: $\frac{df}{f} = \frac{dx}{x}$

Step 6: Integrate both sides of the differential equation. Integrating both sides with respect to their respective variables: $\int \frac{df}{f} = \int \frac{dx}{x}$ $\ln|f(x)| = \ln|x| + \ln C$ Since the domain and codomain are $(0, \infty)$, $f(x) > 0$ and $x > 0$. So we can remove the absolute value signs. $\ln f(x) = \ln x + \ln C$ Using the logarithm property $\ln a + \ln b = \ln(ab)$: $\ln f(x) = \ln(Cx)$ Exponentiating both sides: $f(x) = Cx$ where $C$ is the constant of integration.

Step 7: Use the initial condition to find the value of C. We are given that $f(1) = e$. Substitute $x=1$ and $f(1)=e$ into the equation $f(x) = Cx$: $e = C(1)$ $C = e$ So, the function is $f(x) = ex$.

Step 8: Find the value of x when f(x) = 1. We need to find $x$ such that $f(x) = 1$. Using our derived function $f(x) = ex$: $ex = 1$ Solving for $x$: $x = \frac{1}{e}$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Also, remember to differentiate with respect to the variable approaching the limit.
• Algebraic errors when simplifying: Be careful when factoring and rearranging terms, especially when dealing with function notation and derivatives.
• Forgetting the constant of integration: When solving differential equations, the constant of integration is crucial. The initial condition is used to determine its specific value.
• Handling absolute values: When integrating $\frac{1}{x}$ or $\frac{1}{f(x)}$, remember the $\ln|x|$ or $\ln|f(x)|$. However, in this problem, the domain and codomain restrictions $(0, \infty)$ simplify this by allowing us to drop the absolute value signs.

Summary

The problem involves evaluating a limit that leads to a differential equation. By applying L'Hôpital's Rule to the given limit, we derived the differential equation $f(x) = xf'(x)$. This equation was solved using the method of separation of variables, yielding the general solution $f(x) = Cx$. Using the initial condition $f(1) = e$, we found the constant $C$ to be $e$, giving the specific function $f(x) = ex$. Finally, by setting $f(x) = 1$, we solved for $x$ and found $x = \frac{1}{e}$.

The final answer is $\boxed{\frac{1}{e}}$.