ابتدا باید حاصل داخلی‌ ترین پرانتز را به‌ دست آوریم یعنی $f\left( f\left( x \right) \right)$. برای این کار در تابع $f\left( x \right)$ باید به‌ جای xها مقدار $\frac{{{x}^{2}}}{{{x}^{2}}+1}$ قرار می‌دهیم.

$\begin{align}
  & f\left( x \right)=\frac{{{x}^{2}}}{{{x}^{2}}+1}\to f\left( f\left( x \right) \right)=f\left( \frac{{{x}^{2}}}{{{x}^{2}}+1} \right)=\frac{{{\left( \frac{{{x}^{2}}}{{{x}^{2}}+1} \right)}^{2}}}{{{\left( \frac{{{x}^{2}}}{{{x}^{2}}+1} \right)}^{2}}+1}=\frac{\frac{{{x}^{4}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}{\frac{{{x}^{4}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}+1}=\frac{\frac{{{x}^{4}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}{\frac{{{x}^{4}}+{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}= \\ 
 & \frac{{{x}^{4}}}{{{x}^{4}}+{{\left( {{x}^{2}}+1 \right)}^{2}}}=\frac{{{x}^{4}}}{{{x}^{4}}+{{x}^{4}}+1+2{{x}^{2}}}=\frac{{{x}^{4}}}{2{{x}^{4}}+2{{x}^{2}}+1} \\ 
\end{align}$

$\lim\limits_{x\to 1}\frac{4{{x}^{2}}}{{{x}^{2}}+1}f\left( \left( f\left( x \right) \right) \right)=\lim\limits_{x\to 1}\frac{4{{x}^{2}}}{{{x}^{2}}+1}\times \frac{{{x}^{4}}}{2{{x}^{4}}+2{{x}^{2}}+1}=\frac{4{{\left( 1 \right)}^{2}}}{{{\left( 1 \right)}^{2}}+1}\times \frac{{{\left( 1 \right)}^{4}}}{2{{\left( 1 \right)}^{4}}+2{{\left( 1 \right)}^{2}}+1}=\frac{4}{2}\times \frac{1}{5}=\frac{4}{10}=0/4$