Topics Section 1 6 Calculating Limits Using the Limit Laws Any errors you can nd in the solutions can be reported here and are greatly appreciated https forms gle rGXwBeet5a3c3kF6A 1 Given that lim x 4 f x 6 lim x 4 g x 0 and lim x 4 h x 1 nd the limits below If the limit does not exist explain why a lim x 4 f x 4h x Solution Using limits additive and scaling properties f x 4h x lim x 4 lim x 4 6 4 1 2 lim x 4 4h x f x lim x 4 f x 4 lim x 4 h x Solution We can pull limits inside continuous functions Since g t t3 is a continuous function we have cid 18 cid 19 3 lim x 4 f x 3 lim x 4 f x 63 216 Solution We are allowed to pull limits onto the numerator and denominator since the limit of the denom inator is not zero lim x 4 g x 3h x limx 4 g x limx 4 3h x 0 3 1 0 Solution The limit does not exist since the denominator approaches zero and the numerator does not approach zero 2 Evaluate each limit and justify each step by indicating the appropriate Limit Laws b lim x 4 f x 3 c lim x 4 g x 3h x d lim x 4 h x 2g x a4 8a 4 3a2 16 a lim a 2 Solution We ll be very careful to do each maneuver with limits step by step a4 8a 4 3a2 16 lim a 2 lima 2 a4 8a 4 lima 2 a 4 8 lima 2 a 4 lima 2 3a2 16 3 lima 2 a 2 16 24 8 2 4 3 22 16 0 143 1 cid 114 2u 5 b lim u 1 3u 11 Solution Using limit laws cid 114 2u 5 3u 11 lim u 1 cid 114 cid 115 cid 115 cid 114 3 8 lim u 1 2u 5 3u 11 limu 1 2u 5 limu 1 3u 11 2 1 5 3 1 11 lim x 6 1 x 1 6 x 6 6 x 6x x 6 1 6x 1 6 6 lim x 6 lim x 6 lim x 6 1 36 3 Evaluate the following limit if it exists If the limit does not exist explain why If you use a theorem clearly state which theorem you are using a lim x 6 1 x 1 6 x 6 Solution Let s combine the numerator into one fraction Then the x 6 terms cancel The only limit theorem we use here is that f x 1 6x is a continuous function at x 6 Solution Using polynomial long division another algebra skill to brush up on we nd x3 1 Therefore x 1 x2 x 1 x3 1 x 1 lim x 1 lim x 1 x2 x 1 11 1 1 3 x3 1 x 1 b lim x 1 2v 1 2v 1 c lim v 1 2 Solution Remember that Therefore plugging in 2v 1 and solving the resulting inequality cid 40 x x 0 x x x 0 2v 1 cid 40 cid 40 2v 1 2v 1 2v 1 2v 1 2v 1 0 2v 1 0 v 1 2 v 1 2 Since we are taking the left hand limit in the range x 1 2 we have 2v 1 2v 1 lim v 1 2 lim v 1 2 2v 1 2v 1 1 1 lim v 1 2 2 2v 1 2v 1 d lim v 1 2 Solution Taking the right hand limit we similarly compute 2v 1 2v 1 2v 1 2v 1 lim v 1 lim v 1 2 2 lim v 1 2 1 1 does not exist because the corresponding right hand and left hand limits are not equal Therefore lim v 1 2v 1 2v 1 cid 18 1 cid 19 2 x e lim x 0 x4 cos Solution Using the squeeze theorem cid 18 1 cid 19 x x4 x4 cos x4 Since the red and blue terms have limit 0 as x 0 the squeeze theorem says limx 0 x4 cos 0 2 u2 5 u 3 f lim u 3 Hint multiply by the conjugate Solution We multiply by 1 in the form 1 2 2 then factor and terms cancel cid 18 1 cid 19 x 2 u2 5 lim u 3 u 3 lim u 3 u2 5 u2 5 cid 33 cid 32 u2 5 u2 5 2 2 2 u2 5 u 3 4 u2 5 u 3 2 u 3 u 3 u2 5 u2 5 lim u 3 lim u 3 lim u 3 u 3 2 u 3 u2 5 2 3 3 2 cid 112 3 2 5 Solution Since 1 sin 1 t2 1 and f x 2x is an increasing function Since the red and blue terms approach 0 as t 0 the Squeeze Theorem says that 2 1 2sin 1 t2 21 2 1t2 t22sin 1 t2 21t2 t22sin 1 t2 0 lim t 0 g lim t 0 t22sin 1 t2 and so t 2 3 t 7 h lim t 7 Solution Should be similar to f Your answer should be 1 6 6 4 3 2 3 4 h 2 16 h i lim h 0 cid 18 cos x cid 19 x 4 x 4 2 sin j lim x 4 Solution First expand 4 h 2 16 8h h2 Then do the algebra carefully Your answer should be 8 Solution Answer is zero Similar Squeeze Theorem argument to e and g 4 Is there a number a such that lim x 2 3x2 ax a 3 x2 x 2 exists If so nd the value of a and the value of the limit Solution The denominator factors as x2 x 2 x 2 x 1 For the limit to exist we need x 2 to be a factor in the numerator Why This happens if 2 is a root of the numerator so let s see if we can make that happen Plugging in x 2 into the numerator we want a such that 3 2 2 a 2 a 3 0 12 2a a 3 0 15 a 0 a 15 If we choose a 15 the limit exists Check this yourself try to gure out the limit with a 15 plugged in Factor the numerator and denominator and you should see factors cancel and nd that the limit exists Note We used an important algebra fact here x a is a factor of a polynomial P x P a 0 5 True or False a If lim x 5 f x 0 and lim x 5 g x 0 then lim x 5 f x g x that is an example that satis es the hypothesis but not the conclusion does not exist If the answer is false give a counterexample Solution False Take e g f x x 5 and g …
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