Let $$f : (0, \infty) \to (0, \infty)$$ be a differentiable function such that $$f(1) = e$$ and $$\lim_{t \to x}\frac{t^2f^2(x) - x^2f^2(t)}{t - x} = 0$$. If $$f(x) = 1$$, then $$x$$ is equal to:

We are told that the real-valued function $$f:(0,\infty)\to(0,\infty)$$ is differentiable, that it satisfies $$f(1)=e,$$ and that the limit

$$\lim_{t\to x}\frac{t^{2}f^{2}(x)\;-\;x^{2}f^{2}(t)}{t-x}=0$$

exists and is equal to zero for every positive real number $$x$$ in the domain. We wish to find the point $$x$$ (if any) at which $$f(x)=1$$ holds.

Because the limit involves the difference quotient of an expression that vanishes when $$t=x,$$ we recognise it as the derivative (with respect to $$t$$) of the numerator, evaluated at $$t=x.$$ To formalise this observation we proceed step by step.

Let us define a new auxiliary function

$$k(t)=t^{2}f^{2}(x)-x^{2}f^{2}(t),$$

where $$x>0$$ is fixed and $$t$$ is the variable that approaches $$x.$$ Notice first that

$$k(x)=x^{2}f^{2}(x)-x^{2}f^{2}(x)=0,$$

so both the numerator and the denominator of the given difference quotient vanish when $$t=x,$$ exactly the situation in the usual definition of a derivative. By that definition, for a differentiable function $$k,$$ we have

$$\lim_{t\to x}\frac{k(t)-k(x)}{t-x}=k'(x).$$

Because $$k(x)=0,$$ the given limit becomes simply $$k'(x).$$ Therefore the condition in the problem statement may be rewritten as

$$k'(x)=0 \quad\text{for every } x>0.$$

We now differentiate $$k(t)$$ with respect to its variable $$t.$$ The first term, $$t^{2}f^{2}(x),$$ involves $$t$$ explicitly but contains the constant factor $$f^{2}(x)$$ (constant because $$x$$ is fixed while we differentiate with respect to $$t$$). Using the elementary rule $$\dfrac{d}{dt}\bigl(t^{n}\bigr)=nt^{\,n-1},$$ we obtain

$$\frac{d}{dt}\Bigl[t^{2}f^{2}(x)\Bigr]=2t\,f^{2}(x).$$

The second term, $$-x^{2}f^{2}(t),$$ contains $$t$$ only inside $$f(t).$$ First we differentiate $$f^{2}(t)$$ using the chain rule. Stating the chain rule explicitly: if $$u(t)=f(t),$$ then $$\dfrac{d}{dt}\bigl[u^{2}\bigr]=2\,u\,\dfrac{du}{dt}=2f(t)f'(t).$$ Multiplying by the constant factor $$-x^{2},$$ we get

$$\frac{d}{dt}\Bigl[-x^{2}f^{2}(t)\Bigr]=-x^{2}\cdot 2f(t)f'(t)=-2x^{2}f(t)f'(t).$$

Adding the two differentiated parts, the derivative of $$k(t)$$ is

$$k'(t)=2t\,f^{2}(x)\;-\;2x^{2}f(t)f'(t).$$

We now evaluate this derivative at the point $$t=x$$ (because the limit involves $$t\to x$$):

$$k'(x)=2x\,f^{2}(x)\;-\;2x^{2}f(x)f'(x).$$

According to the earlier deduction, the given limit equalling zero implies

$$k'(x)=0.$$

Hence

$$2x\,f^{2}(x)-2x^{2}f(x)f'(x)=0.$$

We can divide the entire equality by the common positive factor $$2x\;(x>0)$$ to simplify it:

$$f^{2}(x)-x\,f(x)f'(x)=0.$$

Since $$f(x)>0$$ for all $$x$$ in the domain (as the codomain is $$(0,\infty)$$), we may divide once more by the positive quantity $$f(x)$$ to arrive at a very clean first-order differential equation:

$$f(x)-x\,f'(x)=0.$$

We rewrite this as

$$x\,f'(x)=f(x).$$

To make the separation of variables absolutely explicit, we bring all occurrences of $$f$$ to one side and all occurrences of $$x$$ to the other:

$$\frac{f'(x)}{f(x)}=\frac{1}{x}.$$

We now integrate both sides with respect to $$x.$$ Stating the standard integrals: $$\displaystyle\int \frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C_{1}$$ and $$\displaystyle\int\frac{1}{x}\,dx=\ln|x|+C_{2}.$$ Combining the constants into one overall constant $$C,$$ we have

$$\ln f(x)=\ln x + C.$$

Exponentiating both sides to remove the natural logarithm, we obtain

$$f(x)=e^{C}\,x.$$

Let us denote the positive constant $$e^{C}$$ by the single symbol $$K.$$

So the general solution satisfying the differential equation and remaining positive is

$$f(x)=Kx,\qquad K>0.$$

We now employ the initial condition supplied in the problem: $$f(1)=e.$$ Substituting $$x=1$$ into the expression $$f(x)=Kx$$ gives

$$f(1)=K\cdot 1=K.$$

But we are told that $$f(1)=e.$$ Therefore

$$K=e.$$

Consequently the explicit form of our function is

$$f(x)=ex.$$

We are finally ready to answer the specific question, namely to find the positive number $$x$$ that satisfies $$f(x)=1.$$ Substituting the concrete formula for $$f$$ we have derived, we set

$$ex=1.$$

Solving for $$x$$ gives

$$x=\frac{1}{e}.$$

This value coincides with Option A in the list provided.

Hence, the correct answer is Option A.