If $f '$ (0) = 0 and f(x) is a differentiable and increasing function,then lim $x \rightarrow 0$  $\frac {x.f ' (x^2)}{f ' (x)}$

• This is an indeterminate form of type 0/0, so we can apply L'Hôpital's Rule.
  

  $\begin{array}{l}\text { - Numerator: } \frac{d}{d x}\left[x \cdot f^{\prime}\left(x^{2}\right)\right]=f^{\prime}\left(x^{2}\right)+x \cdot f^{\prime \prime}\left(x^{2}\right) \cdot 2 x \\ \text { - Denominator: } \frac{d}{d x}\left[f^{\prime}(x)\right]=f^{\prime \prime}(x)\end{array}$

  
  

  - As $x \rightarrow 0, x^{2} \rightarrow 0$, so $f^{\prime}\left(x^{2}\right) \rightarrow f^{\prime}(0)=0$.

  - The term $x \cdot f^{\prime \prime}\left(x^{2}\right) \cdot 2 x$ also approaches 0 as $x \rightarrow 0$.

  - Therefore, the limit of the numerator's derivative is 0 .

  - The limit of the denominator's derivative is $f^{\prime \prime}(0)$.

  
  

  Since the limit of the numerator's derivative is 0 and the limit of the denominator's derivative is $f^{\prime \prime}(0)$, the overall limit is:

  $\lim _{x \rightarrow 0} \frac{x \cdot f^{\prime}\left(x^{2}\right)}{f^{\prime}(x)}=\frac{0}{f^{\prime \prime}(0)}=0$