Use l'Hôpital's rule to find the limit.
\[
\lim _{x \rightarrow 0} \frac{2 x^{2}}{\cos x-1}
\]
What must be done to put the given expression in a form to which l'Hôpital's rule can be directly applied?

Step 1 ：The given expression is \(\frac{2 x^{2}}{\cos x-1}\). However, this is not in the form of 0/0 or ∞/∞ as x approaches 0, which is the requirement for applying l'Hôpital's rule. We need to manipulate the expression to bring it to this form.

Step 2 ：We can do this by using the trigonometric identity \(\cos(2x) = 2\cos^2(x) - 1\). This will allow us to rewrite the denominator in a way that it approaches 0 as x approaches 0.

Step 3 ：Rewriting the expression, we get \(\frac{2 x^{2}}{\frac{\cos(2x)}{2} - \frac{1}{2}}\).

Step 4 ：Simplifying the expression, we get \(\frac{4 x^{2}}{\cos(2x) - 1}\).

Step 5 ：The expression is now in the form 0/0 as x approaches 0, so we can apply l'Hôpital's rule. l'Hôpital's rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives. So, we need to find the derivative of the numerator and the derivative of the denominator.

Step 6 ：The derivative of the numerator is 8x and the derivative of the denominator is -2sin(2x).

Step 7 ：Applying l'Hôpital's rule, we find that the limit of the given expression as x approaches 0 is -2.

Step 8 ：Final Answer: \(\boxed{-2}\)