$\dfrac {f(x)-f(0)}{g(x)-g(0)}=\dfrac {f'\big( \theta(x)\big)}{g'\big( \theta(x)\big)}$ , $\lim_{x \to 0+} \dfrac{\theta(x)}x=?$

Question

$f,g:[0,1 ]\to [0,1]$ be continuous functions and twice differentiable in $[0,1]$ such that $g'(x) \ne 0 ,\forall x \in (0,1) , f''(0)g'(0) \ne f'(0)g''(0) $ , let $ \theta(x)$ be one of the numbers for which the assertion of the Cauchy's generalized mean-value theorem holds i.e. $\dfrac {f(x)-f(0)}{g(x)-g(0)}=\dfrac {f'\big( \theta(x)\big)}{g'\big( \theta(x)\big)}$ , then how to compute $\lim_{x \to 0+} \dfrac{\theta(x)}x$ ?

Answer 1

Let $x>0$. There exists $c_x,d_x,a_x,b_x \in (0,x)$ such that:

$$f(x)=f(0)+xf^{\prime}(0)+\frac{x^2}{2}f^{\prime\prime}(c_x)$$

$$g(x)=g(0)+xg^{\prime}(0)+\frac{x^2}{2}g^{\prime\prime}(d_x)$$

$$f^{\prime}(\theta(x))=f^{\prime}(0)+\theta(x)f^{\prime\prime}(a_x)$$

and

$$g^{\prime}(\theta(x))=f^{\prime}(0)+\theta(x)f^{\prime\prime}(b_x)$$

Now from your equality, you get:

$$\theta(x)\big [(f^{\prime}(0)g^{\prime\prime}(b_x)-g^{\prime}(0)f^{\prime\prime}(a_x))+\frac{x}{2}(f^{\prime\prime}(c_x)g^{\prime\prime}(b_x)-f^{\prime\prime}(a_x)g^{\prime\prime}(d_x))\big ]=\frac{x}{2}(f^{\prime}(0)g^{\prime\prime}(d_x)-g^{\prime}(0)f^{\prime\prime}(c_x))$$

Now as $x \to 0$:

$$\frac{x}{2}(f^{\prime\prime}(c_x)g^{\prime\prime}(b_x)-f^{\prime\prime}(a_x)g^{\prime\prime}(d_x)) \to 0$$

$$f^{\prime}(0)g^{\prime\prime}(b_x)-g^{\prime}(0)f^{\prime\prime}(a_x)\to f^{\prime}(0)g^{\prime\prime}(0)-g^{\prime}(0)f^{\prime\prime}(0)=L \not =0$$

and $$f^{\prime}(0)g^{\prime\prime}(d_x)-g^{\prime}(0)f^{\prime\prime}(c_x))\to L$$ and so $\displaystyle \frac{\theta(x)}{x}\to \frac{1}{2}$.

Answer 2

More observation than solution: If you let $f(x)=ax^2+bx$ and $g(x)=cx^2+dx$, then, in general, the conditions of the problem are satisfied, and the equation becomes

$${ax^2+bx\over cx^x+dx}={ax+b\over cx+d}={2a\theta+b\over2c\theta+d}$$

which solves to

$${\theta\over x}={1\over2}$$

This accords with the answer just given by Kelenner.

(Note: the OP's comment and my reply were to an earlier version of the answer, which failed to notice the condition $f''(0)g'(0)\not=f'(0)g''(0)$.)