If $f\left(x\right)$ is a differentiable function such that $f^{\text{'}}\left(1\right)=4$ and $f^{\text{'}}\left(4\right)=\frac{1}{2},$ then value of $\underset{x\to 0}{\lim }\frac{f(x^2+x+1)-f\left(1\right)}{f(x^4-x^2+2x+4)-f\left(4\right)}$ is

$\underset{x\to 0}{\lim }\frac{f(x^2+x+1)}{f(x^4-x^2+2x+4)-f\left(4\right)}\left(\frac{0}{0}form\right)$

As. f (x) is differentiable then apply $L\text{'}$ Hospital rule

$\underset{x\to 0}{\lim }\frac{(2x+1)f\text{'}(x^2+x+1)}{(4x^3-2x+2)f\text{'}(x^4-x^2+2x+4)}=\frac{f\text{'}\left(1\right)}{2f\text{'}\left(4\right)}=4$