If $$f(x)$$ is a differentiable function in the interval $$(0, \infty)$$ such that $$f(1) = 1$$ and $$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$$, for each $$x \gt 0$$, then $$f\left(\frac{3}{2}\right)$$ is equal to

We are told that the real-valued function $$f(x)$$ is differentiable for every $$x\gt 0$$, that it satisfies the initial condition $$f(1)=1$$ and, in addition, that for every $$x\gt 0$$ the following limit holds

$$\lim_{t\to x}\dfrac{t^{2}f(x)-x^{2}f(t)}{t-x}=1.$$

Because the limit is of the standard “difference quotient” type, we first rewrite the numerator so that a familiar derivative expression appears. We have

$$t^{2}f(x)-x^{2}f(t) =\bigl[t^{2}f(x)-x^{2}f(x)\bigr]-x^{2}\bigl[f(t)-f(x)\bigr] =(t^{2}-x^{2})f(x)-x^{2}\bigl[f(t)-f(x)\bigr].$$

Notice that $$t^{2}-x^{2}=(t-x)(t+x)$$, so dividing the whole numerator by $$t-x$$ gives

$$\dfrac{t^{2}f(x)-x^{2}f(t)}{t-x} =(t+x)f(x)-x^{2}\dfrac{f(t)-f(x)}{t-x}.$$

Now let $$t\to x$$. Using the continuity of addition and the definition of the derivative,

$$\lim_{t\to x}(t+x)=2x, \qquad \lim_{t\to x}\dfrac{f(t)-f(x)}{t-x}=f'(x).$$

Therefore the given limit statement becomes

$$2x\,f(x)-x^{2}\,f'(x)=1\quad\text{for every }x\gt 0.$$

We have obtained a first-order linear differential equation:

$$-x^{2}f'(x)+2x\,f(x)=1.$$ Multiplying by $$-1$$ to get the standard “$$f'$$ + ( something ) $$f$$” form, we find

$$x^{2}f'(x)-2x\,f(x)=-1,$$ or, after dividing by $$x^{2}$$ (which is always positive in the interval $$(0,\infty)$$),

$$f'(x)-\dfrac{2}{x}\,f(x)=-\dfrac{1}{x^{2}}.$$

This is a linear ordinary differential equation of the form $$f'(x)+P(x)f(x)=Q(x)$$ with

$$P(x)=-\dfrac{2}{x},\qquad Q(x)=-\dfrac{1}{x^{2}}.$$

First we find the integrating factor. By definition, the integrating factor $$\mu(x)$$ satisfies

$$\mu(x)=\exp\!\Bigl(\int P(x)\,dx\Bigr) =\exp\!\Bigl(\int -\dfrac{2}{x}\,dx\Bigr) =\exp(-2\ln x)=x^{-2}.$$

Multiplying the entire differential equation by this integrating factor gives

$$x^{-2}f'(x)-\dfrac{2}{x}\,x^{-2}f(x)=-x^{-4}.$$ But the left-hand side is exactly the derivative of $$x^{-2}f(x)$$, because

$$\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr] =x^{-2}f'(x)+(-2x^{-3})f(x) =x^{-2}f'(x)-\dfrac{2}{x}\,x^{-2}f(x).$$

Hence we may rewrite the equation compactly as

$$\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr]=-x^{-4}.$$

We now integrate both sides with respect to $$x$$:

$$\int\dfrac{d}{dx}\bigl[x^{-2}f(x)\bigr]\,dx =\int -x^{-4}\,dx.$$

The left integral simply returns the inside expression, while the right integral is computed using the power rule $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C$$:

$$x^{-2}f(x) =-\dfrac{x^{-3}}{-3}+C =\dfrac{1}{3}x^{-3}+C,$$ where $$C$$ is the constant of integration.

Finally we multiply by $$x^{2}$$ to solve for $$f(x)$$ explicitly:

$$f(x)=\dfrac{1}{3}x^{-1}+C\,x^{2}.$$

The initial condition $$f(1)=1$$ determines $$C$$. Substituting $$x=1$$ gives

$$1=f(1)=\dfrac{1}{3}(1)^{-1}+C(1)^{2} =\dfrac{1}{3}+C,$$ so

$$C=1-\dfrac{1}{3}=\dfrac{2}{3}.$$

Therefore the required function is

$$f(x)=\dfrac{2}{3}x^{2}+\dfrac{1}{3x}.$$

Now we must evaluate $$f\!\left(\dfrac{3}{2}\right)$$. Substituting $$x=\dfrac{3}{2}$$ yields

$$f\!\left(\dfrac{3}{2}\right) =\dfrac{2}{3}\left(\dfrac{3}{2}\right)^{2} +\dfrac{1}{3\left(\dfrac{3}{2}\right)} =\dfrac{2}{3}\cdot\dfrac{9}{4} +\dfrac{1}{\dfrac{9}{2}} =\dfrac{18}{12}+\dfrac{2}{9} =\dfrac{3}{2}+\dfrac{2}{9}.$$

To add the two fractions, convert both to the common denominator $$18$$:

$$\dfrac{3}{2}=\dfrac{27}{18},\qquad \dfrac{2}{9}=\dfrac{4}{18},$$

so

$$f\!\left(\dfrac{3}{2}\right) =\dfrac{27}{18}+\dfrac{4}{18} =\dfrac{31}{18}.$$

Hence, the correct answer is Option D.