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1/4 2/4 3/4 4/4 ㅠ cot² (x) 0 Substituting x = into results in the indeterminate form This doesn't 2 1 - sin(x) 0 necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have cos² (x) in our expression, because cot² (x) = the Pythagorean identity, sin²(x) + cos²(x) = 1: cot²(x) 1 - sin(x) cos²(x) sin²(x) [1 - sin(x)] (1 — sin²(x)) sin²(x) [1 - sin(x)] (1 sin(x))(1+ sin(x)) sin²(x) [1 - sin(x)] 1+ sin(x) sin²(x) (1-sin(x))(1+ sin(x)) sin²(x) [1-sin(x)] except for (2k + 1) In our case, x = Therefore lim π 2 2 for x {...,- " Definition of cotangent This means that the two expressions have the same value for all x-values (in their domains) ㅠ ㅠ -1)/1 for any integer k, and specifically 2 2 cot² (x) 1 - sin(x) The Pythagorean identity cot²(x) 1-sin(x) We can now use the following theorem: If f(x) = g(x) for all x-values in a given interval except for x = c, then lim f(x) = lim g(x). x→C x-c lim In conclusion, lim 2 7π 2 Diff. of squares Cancel common factors Sin 2 1 + sin(x) ₁²(x) 3п п 5п 9п 22 2¹ 2 1+ sin(x) sin²(x) = 2. cos² (x) sin(x) ,...} (The last limit was found using direct substitution.) cot² (x) T1 - sin(x) 2 = 2 -, let's rewrite it using for all x-values in the interval (-π, π) except for