The value of f (0) such that the function f(x) = (1 + 2x3 - 1 +...

For $f\left(x\right)$ to be continuous at $x = 0,$ we must have
$f\left(0\right) = \lim_{x \to 0}\dfrac{1 + 2x3 - 1 + x4}{x}$
$= \lim_{x \to 0}\dfrac{\left[1 + \dfrac{2}{3}x + \dfrac{\left(\dfrac{1}{3}\right)\left(- \dfrac{2}{3}\right)}{2 !}4{x}^{2} + . . . . .\right]}{x} - \dfrac{\left[1 + \dfrac{1}{4}x + \dfrac{\left(\dfrac{1}{4}\right)\left(- \dfrac{3}{4}\right)}{2 !}{x}^{2} + . . . . .\right]}{x}$
$= \lim_{x \to 0}\dfrac{x\left[\dfrac{5}{12} + \left(\dfrac{- 4}{9} + \dfrac{3}{32}\right)x + termscontaining{x}^{2}andhigherpowers\right]}{x}$
$= \dfrac{5}{12}$

$∴ f\left(0\right) = \dfrac{5}{12}$