Question

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

Answer 1

$\underset{x \rightarrow 0}{lim}\frac{f \left(x^{2} + x + 1\right) - f \left(\right. 1 \left.\right)}{f \left(x^{4} - x^{2} + 2 x + 4\right) - f \left(\right. 4 \left.\right)}$ form $\left.$ $\left(\frac{0}{0} form\right)$
$\underset{x \rightarrow 0}{lim}\frac{\left(\right. 2 x + 1 \left.\right) f^{1} \left(x^{2} + x + 1\right)}{\left(4 x^{3} - 2 x + 2\right) f^{1} \left(x^{4} - x^{2} + 2 x + 4\right)}$
$=\frac{f^{'} \left(\right. 1 \left.\right)}{2 f^{'} \left(\right. 4 \left.\right)}=4$
(Applying $L$ ' Hospital's Rule)