Key Concepts and Formulas

• Definition of the Derivative: The derivative of a function $f$ at a point $x$, denoted by $f'(x)$, can be defined using a limit: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. A related form is $\lim_{t \to x} \frac{f(t) - f(x)}{t-x} = f'(x)$.
• L'Hôpital's Rule: For limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, if $\lim_{x \to c} \frac{g(x)}{h(x)}$ is an indeterminate form, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• First-Order Linear Differential Equations: An equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ is a first-order linear differential equation. The integrating factor (I.F.) is given by $e^{\int P(x) dx}$. The general solution is then $y \cdot \text{I.F.} = \int Q(x) \cdot \text{I.F.} dx + C$.

Step-by-Step Solution

Step 1: Analyze the given limit and apply L'Hôpital's Rule. We are given the limit: $\lim_{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ As $t \rightarrow x$, the numerator approaches $x^2 f(x) - x^2 f(x) = 0$, and the denominator approaches $x-x=0$. This is an indeterminate form of type $\frac{0}{0}$. Therefore, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $t$ (treating $x$ as a constant).

Differentiating the numerator with respect to $t$: $\frac{d}{dt}(t^2 f(x) - x^2 f(t)) = 2t f(x) - x^2 f'(t)$. Differentiating the denominator with respect to $t$: $\frac{d}{dt}(t-x) = 1$.

Applying L'Hôpital's Rule: $\lim_{t \rightarrow x} \frac{2t f(x) - x^2 f'(t)}{1} = 1$

Step 2: Evaluate the limit after applying L'Hôpital's Rule to obtain a differential equation. Now, substitute $t=x$ into the expression obtained in Step 1: $2x f(x) - x^2 f'(x) = 1$ This equation relates the function $f(x)$ and its derivative $f'(x)$. We can rewrite this as a differential equation. Let $y = f(x)$, so $y' = f'(x)$. $2x y - x^2 y' = 1$ Rearranging this to match the standard form of a first-order linear differential equation ($\frac{dy}{dx} + P(x)y = Q(x)$): $-x^2 y' + 2x y = 1$ Divide by $-x^2$ (since $x > 0$, $x^2 \neq 0$): $y' - \frac{2x}{x^2} y = -\frac{1}{x^2}$ $y' - \frac{2}{x} y = -\frac{1}{x^2}$ This is a first-order linear differential equation with $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.

Step 3: Calculate the integrating factor (I.F.). The integrating factor is given by $e^{\int P(x) dx}$. $\text{I.F.} = e^{\int -\frac{2}{x} dx}$ $\text{I.F.} = e^{-2 \ln|x|}$ Since the domain is $(0, \infty)$, $x > 0$, so $|x| = x$. $\text{I.F.} = e^{-2 \ln x} = e^{\ln x^{-2}} = x^{-2} = \frac{1}{x^2}$

Step 4: Solve the differential equation using the integrating factor. Multiply the differential equation by the integrating factor: $\frac{1}{x^2} \left( y' - \frac{2}{x} y \right) = \frac{1}{x^2} \left( -\frac{1}{x^2} \right)$ $\frac{1}{x^2} y' - \frac{2}{x^3} y = -\frac{1}{x^4}$ The left side is the derivative of the product of $y$ and the integrating factor: $\frac{d}{dx} \left( y \cdot \frac{1}{x^2} \right) = -\frac{1}{x^4}$ Now, integrate both sides with respect to $x$: $\int \frac{d}{dx} \left( \frac{y}{x^2} \right) dx = \int -\frac{1}{x^4} dx$ $\frac{y}{x^2} = -\int x^{-4} dx$ $\frac{y}{x^2} = - \frac{x^{-3}}{-3} + C$ $\frac{y}{x^2} = \frac{1}{3x^3} + C$ where $C$ is the constant of integration.

Step 5: Use the initial condition $f(1)=1$ to find the value of $C$. We are given that $f(1)=1$. This means when $x=1$, $y=1$. Substitute these values into the equation from Step 4: $\frac{1}{1^2} = \frac{1}{3(1)^3} + C$ $1 = \frac{1}{3} + C$ $C = 1 - \frac{1}{3} = \frac{2}{3}$

Step 6: Determine the function $f(x)$. Substitute the value of $C$ back into the equation from Step 4: $\frac{y}{x^2} = \frac{1}{3x^3} + \frac{2}{3}$ Now, solve for $y$ (which is $f(x)$): $y = x^2 \left( \frac{1}{3x^3} + \frac{2}{3} \right)$ $y = \frac{x^2}{3x^3} + \frac{2x^2}{3}$ $y = \frac{1}{3x} + \frac{2x^2}{3}$ So, the function is $f(x) = \frac{1}{3x} + \frac{2x^2}{3}$.

Step 7: Calculate $2f(2) + 3f(3)$. First, find $f(2)$: $f(2) = \frac{1}{3(2)} + \frac{2(2)^2}{3} = \frac{1}{6} + \frac{2(4)}{3} = \frac{1}{6} + \frac{8}{3} = \frac{1}{6} + \frac{16}{6} = \frac{17}{6}$ Next, find $f(3)$: $f(3) = \frac{1}{3(3)} + \frac{2(3)^2}{3} = \frac{1}{9} + \frac{2(9)}{3} = \frac{1}{9} + \frac{18}{3} = \frac{1}{9} + 6 = \frac{1}{9} + \frac{54}{9} = \frac{55}{9}$ Now, calculate $2f(2) + 3f(3)$: $2 f(2) + 3 f(3) = 2 \left( \frac{17}{6} \right) + 3 \left( \frac{55}{9} \right)$ $= \frac{34}{6} + \frac{165}{9}$ Simplify the fractions: $= \frac{17}{3} + \frac{55}{3}$ $= \frac{17+55}{3} = \frac{72}{3}$ $= 24$

Common Mistakes & Tips

• Incorrect application of L'Hôpital's Rule: Ensure that the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Also, remember to differentiate with respect to the variable that is approaching the limit ($t$ in this case), treating the other variable ($x$) as a constant.
• Algebraic errors in solving the differential equation: Pay close attention to signs, exponents, and fractions when calculating the integrating factor and integrating. A small error can lead to a completely incorrect function.
• Mistakes in substituting the initial condition: Double-check the values of $x$ and $y$ when using the initial condition to find the constant of integration. Ensure you are using the correct initial condition ($f(1)=1$).

Summary

The problem involves a limit definition of a derivative which, after applying L'Hôpital's Rule, leads to a first-order linear differential equation. By solving this differential equation using an integrating factor and the given initial condition $f(1)=1$, we find the explicit form of the function $f(x)$. Finally, we evaluate the expression $2f(2) + 3f(3)$ using the derived function.

The final answer is $\boxed{24}$.