2 - 2x Let's compute the lim x-2 sin (x - 2) Let's see what happens if we take the limit of the numerator and denominator: lim (a2 - 2x) = 0 x -+ 2 lim (sin(x - 2)) = 0 2 - 2 So we have a limit of indeterminant type O none of the above , so we will try to use L'H rule. ac - 2x (ac2 - 2ac) So instead of working with lim we can take the limit of lim where x-+2 sin (x - 2) x-+2 (sin(x - 2))' (202 - 220) ' = 2ac - 2 (sin(x - 2))' = cos (a - 2) (202 - 2ac) ' By computing the limit we get that lim x -2 (sin(x - 2))' Thus by L'H Rule we get that lim x-2 sin (x - 2)In(x - 2) Let's compute the lim x -+ 00 x2 - 3x Let's see what happens if we take the limit of the numerator and denominator: lim (In(x - 2)) = 00 x -+ 00 lim (202 - 32) 00 So we have a limit of indeterminant type ol c O 0O O none of the above so we will try to use L'H rule. In(x - 2) So instead of working with lim (In(x - 2)) we can take the limit of lim where x + 00 x2 - 3x 20 -+ 00 (ac2 - 3ac) (In(x - 2))' x - 2 (ac2 - 320) ' = 2x - 3 (In(x - 2))' By computing the limit we get that lim 20 -+ 00 (a2 - 3ac) ' In(x - 2) Thus by L'H Rule we get that lim 2-+ 00 x2 - 3xarctan(x) A student is trying to compute lim +1 1 H arctan(x) x2 +1 They try to use L'H rule to get that lim x2 + 1 is equal to lim They can rewrite x -+ 00 + 1 1 1 x2 2-2 as They end up getting that lim = - 1. xc2 + 1 a -+ 00 x2 + 1 arctan(x) Therefore by L'H rule, lim = - 1 x -+00 +1 arctan(x) However, when they graphed and saw what happened as x -> oo they saw that the answer is +1 not - 1. What did they do wrong? Select all that apply.. . arctan(:z:) ' _ C] the student took the derivative of incorrectly. They needed to use quotient rule because i + 1 L'H says you take the derivative of the entire function C] the student took the derivative of arctan(m) incorrectly 1 [:1 the student took the derivative of + 1 incorrectly :1: 1 mgll _ C] the student rewrote 1 Incorrectly _ E? m2 C] the student took the limit of lim incorrectly 1:000 332 + 1 _ arctan(:u) _ _ ' the student cannot use L'H because 11m 1 15 not of indeterminate type $000 _ + 1 m C] the student did nothing wrong; they simply misread the graph What is the actual answer? (Hint Chapter 1.6 may be useful) S