Evaluating a 0/0 limit using repeated L'Hopital's Rule

First confirm the form: as $x \to 0$, the numerator $\cos(x) - 1 + x^2/2 \to 1 - 1 + 0 = 0$ and the denominator $x^4 \to 0$, so this is the indeterminate form $0/0$. Differentiating numerator and denominator separately gives $\frac{-\sin(x) + x}{4x^3}$. Substituting $x=0$ again gives $0/0$, so L'Hopital's Rule applies again.

Differentiate again to get $\frac{-\cos(x) + 1}{12x^2}$, which is still $0/0$ at $x=0$, so apply L'Hopital's Rule a third time. This yields $\frac{\sin(x)}{24x}$, which is again $0/0$; applying the rule once more gives $\frac{\cos(x)}{24}$. Now substituting $x=0$ gives $\frac{1}{24}$, which is not indeterminate, so this is the value of the original limit.