1. Evaluate the limit or determine that it does not exist. (Not sure if I did this correctly or how to do it if I'm wrong) 2. lim(x⇒π/2) tan^2(x) ln(sin(x)) -In text form: the limit as x goes to pi over two, of tangent squared of x, times the natural log of sine of x. 3. My attempt was to change tan^2[x] into sin^2[x]/cos^2[x], where the limit would then give me 0/0, I then used L'Hospital's rule to get the numerator (not sure if I did the derivative correctly with the product rule): 2cos[2x]ln[sin(x)] + -csc^2[x]sin^2[x] and the denominator as -2cos[2x] This gave me -1/0 which would mean the limit is zero, and does exist.