Let  $f\left(x\right)=x^3$  use mean value theorem to write $\frac{f\left(x+h\right)-f\left(x\right)}{h}=f^1\left(x+\theta h\right)$  with  $0<\theta <1.$ If  $x\ne 0$ then $\underset{h\to 0}{\lim }\theta$  is equal to

Given, $f\left(x\right)=x^3$

$\therefore f\left(x+h\right)={\left(x+h\right)}^3$

Now, $f^1\left(x\right)=3x^2$

$\therefore f^1\left(x+\theta h\right)=3{\left(x+\theta h\right)}^2$

Given, $\frac{f\left(x+h\right)-f\left(x\right)}{h}=f^1\left(x+\theta h\right)$

$\Rightarrow \frac{{\left(x+h\right)}^3-x^3}{h}=3{\left(x+\theta h\right)}^2$

$\Rightarrow \frac{x^3+h^3+3xh\left(x+h\right)-x^3}{h}=3\left(x^2+{\theta }^2{h}^2+2x\theta h\right)$

$\Rightarrow h^2+3x^2+3xh=3x^2+3{\theta }^2{h}^2+6x\theta hA$

$\Rightarrow h+3x=3{\theta }^{2}h+6x\theta$

Taking limit on both sides, we get 

$\underset{h\to 0}{\lim }\left(h+3x\right)=\underset{h\to 0}{\lim }\left(3{\theta }^{2}h+6x\theta \right)$

$\Rightarrow 3x=6x\underset{h\to 0}{\lim }\theta =\frac{1}{2}=0.5$