The value of f (0) so that the function f (x) = 1 - (1 - x) x^ 4 is...

If $f$\left(x\right)$is continuous at$x = 0$<br /><br />$f\left(x\right)= \lim_{\text{x} \to 0} \text{f} \left(\text{x}\right)$<br /><br />$= \lim_{\text{x} \to 0} \dfrac{1 - \text\cos \left(1 - \text\cos \text{x}\right)}{{\text{x}}^{4}} \times \dfrac{1 + \text\cos \left(1 - \text\cos \text{x}\right)}{1 + \text\cos \left(1 - \text\cos \text{x}\right)}$<br /><br />$= \lim_{\text{x} \to 0} \dfrac{{\text\sin}^{2} \left(1 - \text\cos \text{x}\right)}{{\text{x}}^{4} \cdot \left(1 + \text\cos \left(1 - \text\cos \text{x}\right)\right)} \cdot \dfrac{{\left(1 - \text\cos \text{x}\right)}^{2}}{{\left(1 - \text\cos \text{x}\right)}^{2}}$<br /><br />$= \lim_{\text{x} \to 0} {\left[\dfrac{\text\sin \left(1 - \text\cos \text{x}\right)}{\left(1 - \text\cos \text{x}\right)}\right]}^{2} \times \lim_{\text{x} \to 0} {\left(\dfrac{1 - \text\cos \text{x}}{{\text{x}}^{2}}\right)}^{2} \times \lim_{\text{x} \to 0} \dfrac{1}{1 + \text\cos \left(1 - \text\cos \text{x}\right)}$<br /><br />$= {\left(1\right)}^{2} \times \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}$<br /><br />$∴ \text{k} = \dfrac{1}{8} ⇒ 10\text{k} = \dfrac{5}{4} = 1.25