Use l'Hôpital's Rule to find the following limit. \[ \lim _{x \rightarrow 0} \frac{5 x^{2}}{2 \ln (\sec x)^{t}} \]

Step 1 ：First, we need to check if the limit is in the form of 0/0 or ∞/∞ to apply l'Hôpital's Rule. As x approaches 0, the numerator \(5x^2\) approaches 0. However, \(\ln(\sec(x))^t\) is undefined at x=0, so we cannot directly apply l'Hôpital's Rule.

Step 2 ：To solve this, we can rewrite the denominator using the properties of logarithms. We have \(\ln(\sec(x))^t = t \ln(\sec(x))\).

Step 3 ：Now, as x approaches 0, \(\sec(x)\) approaches 1, and \(\ln(1) = 0\). So, the denominator \(t \ln(\sec(x))\) also approaches 0. Therefore, the limit is in the form of 0/0, and we can apply l'Hôpital's Rule.

Step 4 ：Applying l'Hôpital's Rule, we take the derivative of the numerator and the denominator. The derivative of \(5x^2\) is \(10x\), and the derivative of \(t \ln(\sec(x))\) is \(t \sec(x) \tan(x)\).

Step 5 ：So, the limit becomes \(\lim_{x \rightarrow 0} \frac{10x}{t \sec(x) \tan(x)}\).

Step 6 ：As x approaches 0, the numerator \(10x\) approaches 0, and the denominator \(t \sec(x) \tan(x)\) approaches 0. Therefore, the limit is still in the form of 0/0, and we can apply l'Hôpital's Rule again.

Step 7 ：Taking the derivative of the numerator and the denominator again, we get \(\lim_{x \rightarrow 0} \frac{10}{t (\sec^2(x) \tan(x) + \sec(x) \sec^2(x))}\).

Step 8 ：As x approaches 0, the numerator is 10, and the denominator \(t (\sec^2(x) \tan(x) + \sec(x) \sec^2(x))\) approaches 0. Therefore, the limit is in the form of ∞/0, which is undefined.

Step 9 ：Therefore, the original limit \(\lim_{x \rightarrow 0} \frac{5x^2}{2 \ln(\sec(x))^t}\) does not exist.