Question 1

Given the limit equation: $\lim_{t \to x} \frac{t^{10} f(x) - x^{10} f(t)}{t^9 - x^9} = 1$ As $t \to x$, this is a $\frac{0}{0}$ form. Applying L'Hopital's Rule with respect to $t$: $\lim_{t \to x} \frac{10t^9 f(x) - x^{10} f'(t)}{9t^8} = 1$ Substituting $t = x$: $\frac{10x^9 f(x) - x^{10} f'(x)}{9x^8} = 1 \implies \frac{10x f(x) - x^2 f'(x)}{9} = 1$ $10x f(x) - x^2 f'(x) = 9 \implies x^2 f'(x) - 10x f(x) = -9$ Dividing by $x^2$: $f'(x) - \frac{10}{x} f(x) = -\frac{9}{x^2}$ This is a linear differential equation. The integrating factor is $I.F. = e^{\int -\frac{10}{x} dx} = e^{-10 \ln x} = x^{-10} = \frac{1}{x^{10}}$. Multiplying by $I.F.$ and integrating: $f(x) \cdot \frac{1}{x^{10}} = \int -\frac{9}{x^2} \cdot \frac{1}{x^{10}} dx = \int -9x^{-12} dx = \frac{9}{11}x^{-11} + C$ $f(x) = \frac{9}{11x} + Cx^{10}$ Using the initial condition $f(1) = 2$: $2 = \frac{9}{11} + C(1)^{10} \implies C = 2 - \frac{9}{11} = \frac{13}{11}$ Thus, $f(x) = \frac{9}{11x} + \frac{13}{11}x^{10}$.