Let's begin by understanding the problem. We have a continuously differentiable function f of x on the interval from 0 to infinity. We're given that f of 1 equals 2, and there's a limit condition that equals 1 for each x greater than 0. Our goal is to find the explicit form of f of x.The given limit has the indeterminate form zero over zero as t approaches x. We can apply L'Hopital's rule by taking the derivative of both numerator and denominator with respect to t. The derivative of t to the tenth times f of x is 10 t to the ninth times f of x, since f of x is constant with respect to t. The derivative of x to the tenth times f of t is x to the tenth times f prime of t. The derivative of t to the ninth is 9 t to the eighth.Now we substitute t equals x into the derivative expression. This gives us 10 x to the ninth times f of x minus x to the tenth times f prime of x, all divided by 9 x to the eighth, equals 1. Multiplying both sides by 9 x to the eighth, we obtain the key differential equation: 10 x to the ninth times f of x minus x to the tenth times f prime of x equals 9 x to the eighth.Let's divide the entire equation by x to the tenth to simplify. This gives us f prime of x equals 10 f of x over x minus 9 over x squared. Rearranging, we get f prime of x minus 10 over x times f of x equals negative 9 over x squared. This is a first-order linear ordinary differential equation in standard form, where P of x equals negative 10 over x and Q of x equals negative 9 over x squared.To solve this first-order linear ODE, we need an integrating factor. The integrating factor mu of x is e to the integral of P of x dx. Substituting P of x equals negative 10 over x, we get e to the integral of negative 10 over x dx, which equals e to the negative 10 natural log of x. Using logarithm properties, this simplifies to e to the natural log of x to the negative tenth, which equals x to the negative tenth. Now we multiply the entire differential equation by this integrating factor.After multiplying by the integrating factor, we get x to the negative tenth times f prime of x minus 10 x to the negative eleventh times f of x equals negative 9 x to the negative twelfth. The left side is exactly the derivative of x to the negative tenth times f of x, by the product rule. The derivative of x to the negative tenth times f of x equals x to the negative tenth times f prime of x plus f of x times negative 10 x to the negative eleventh, which simplifies to our left side. Therefore, we can write d dx of x to the negative tenth times f of x equals negative 9 x to the negative twelfth.Now we integrate both sides with respect to x. The integral of d dx of x to the negative tenth times f of x is simply x to the negative tenth times f of x. The integral of negative 9 x to the negative twelfth is negative 9 times x to the negative eleventh divided by negative 11, plus a constant C. This simplifies to 9 over 11 times x to the negative eleventh plus C. So we have x to the negative tenth times f of x equals 9 over 11 times x to the negative eleventh plus C.To solve for f of x, we multiply both sides by x to the tenth. This gives us f of x equals x to the tenth times the quantity 9 over 11 times x to the negative eleventh plus C. Distributing x to the tenth, we get f of x equals 9 over 11 times x to the negative 1 plus C times x to the tenth. This simplifies to f of x equals 9 over 11 x plus C times x to the tenth. This is the general solution, where C is a constant that we need to determine using the initial condition.Now we apply the initial condition f of 1 equals 2 to find the constant C. Substituting x equals 1 into our general solution, we get f of 1 equals 9 over 11 times 1 plus C times 1 to the tenth, which equals 2. This simplifies to 9 over 11 plus C equals 2. Solving for C, we get C equals 2 minus 9 over 11. Converting 2 to elevenths, we have 22 over 11 minus 9 over 11, which equals 13 over 11. Therefore, C equals 13 over 11.Finally, we substitute C equals 13 over 11 back into the general solution. This gives us f of x equals 9 over 11 x plus 13 over 11 times x to the tenth. Comparing this with the given options, we see that this exactly matches option D. Therefore, the answer is option D: f of x equals 9 over 11 x plus 13 over 11 times x to the tenth.