What is the limit for this function as x approaches 0? y(x)=(cos x)^(1/sin x)

When we try to simply compute y(0) we get 1 to infinity, which is indeterminate (we means we can either get 0, 1, infinity or God knows what else!). The way we want to approach these types of exercises is to play with the functions in order to create a fraction, so we can apply L'Hopital's Rule to the new limit! (a very useful rule which I can easily prove if the reader requires me to do so)By taking a step back, we notice that y(x) can be written as: f(x)^(1/g(x)), where f(x)=cos x and g(x)=sin x. (The reason I make this notation is simple. This trick is a very common one and it is useful for students to recognize it early on)Now we are going to use the wonders of natural logarithms! f(x)^(1/g(x)) is the same thing as e^[ln(f(x))/g(x)]. So our new mission is to find the limit of the exponential, which is now a fraction! By applying L'Hopital's Rule we get:Limit as x approaches 0 of ln(cos x)/sin x is equal to ...BE VERY CAREFUL WHEN APPLYING THE CHAIN RULE!!! :)... the limit as x approaches 0 of (1/cos x) * (-sin x) / cos x which is 0.Thus our initial limit is simply e to the power of 0, which gives us 1.