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**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.

-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.

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.