Find the following limit. Is the function continuous at the point being approached? \[ \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{18-2 \sec t}}\right) \]

Calculate the limit \( \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{18-2 \sec t}}\right) \). Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{t\rightarrow 0}\left(\cos\left(\frac{\pi }{\sqrt{18-2\sec\left(t\right)}}\right)\right)\) - step1: Rewrite the expression: \(\cos\left(\lim _{t\rightarrow 0}\left(\frac{\pi }{\sqrt{18-2\sec\left(t\right)}}\right)\right)\) - step2: Calculate: \(\cos\left(\frac{\pi }{4}\right)\) - step3: Calculate the trigonometric value: \(\frac{\sqrt{2}}{2}\) The limit of the function as \( t \) approaches 0 is \( \frac{\sqrt{2}}{2} \). To determine if the function is continuous at the point being approached, we need to check if the limit exists and is equal to the value of the function at that point. Since the limit exists and is equal to \( \frac{\sqrt{2}}{2} \), the function is continuous at the point being approached.