Math 221 Section 1 5 The Limit of a Function Solutions Courtesy of Jane Davis Any errors you can find in the solutions can be reported here and are greatly appreciated https forms gle rGXwBeet5a3c3kF6A 1 Make an educated guess of the value of the following limits s 3 u2 cos u a lim s 5 lim u 2 b c lim v 4 v 3 4v 2 Solution These are all continuous functions Functions a and b are continuous everywhere the functions graphs never have any jumps or blow ups The third function is continuous except where it s not defined when 4v 2 0 which happens when v 1 2 This is not a problem here because we re taking the limit as v moves towards 4 So in each case the limit is the same as the value of the function with s 5 u 2 v 4 plugged in respectively s 3 5 3 2 u2 cos u 2 2 cos 2 4 1 3 a lim s 5 lim u 2 b c lim v 4 v 3 4v 2 4 3 4 4 2 7 14 1 2 2 Sketch the graph of an example of a function f that satisfies all of the following lim x 3 f x 2 x 1 f x 4 lim lim x 1 f x 1 f 3 4 f 1 1 lim x 3 f x 2 Solution Here s an example The blue circles represent holes and the red circles represent function values 1 3 Determine the infinite limit a lim s 1 s2 4 s 1 b lim u 3 u2 2u 8 u2 6u 9 Solution The numerator approaches 1 2 4 3 The denominator approaches 0 from the left hand side where the function g s s 1 is negative Therefore the limit is positive infinity a negative divided by a negative is positive s2 4 s 1 lim s 1 Solution Factoring the numerator and denominator is usually a good idea u2 2u 8 u2 6u 9 lim u 3 u 4 u 2 u 3 2 lim u 3 Now we can see that the numerator approaches 3 4 3 2 5 as u moves towards 3 from the right and the denominator approaches 0 and is always positive a square is always positive Hence the limit is negative infinity negative divided by positive is negative u2 2u 8 u2 6u 9 lim u 3 t c lim t 9 t 9 3 d lim 4 sin Solution The numerator approaches values t 9 is negative to the left of t 9 and therefore so is t 9 3 So the limit is 9 3 and the denominator approaches 0 through negative Solution Similar the answer is At we have 4 4 0 and sin is negative when is near on the right hand side draw the graph A negative divided by a negative is positive 2 4 Consider the function f x 2x 3 x 2 x 4 a Find all the vertical asymptotes of f Solution The denominator is zero when x 2 and x 4 The numerator is zero when x 3 2 which does not duplicate any of the denominator s zeros Therefore there are vertical asymptotes at x 2 and x 4 b Compute lim x 2 f x lim x 2 f x lim x 4 f x and lim x 4 f x Solution They are based on analyzing which terms are positive or negative as we move from the left hand or right hand side c Make a rough sketch of the function Solution Red lines represent vertical asymptotes 5 Consider the functions f x x 2 g x Sketch each of the functions Then determine the limit as x 3 of each of the functions If the limit does not exist state so and h x x 3 8 x 3 x 2 x 3 x 2 x 3 x 3 x 3 limx 3 f x 3 2 5 since f x is continuous everywhere We find that g x simplifies to Solution g x x 2 with a hole at x 3 So we can find the limit limx 3 g x 3 2 5 Likewise we ignore the jump of h x at x 3 and find limx 3 h x 3 2 5 3 6 Below is the graph of g t For each of the given points determine the value of g a lim t a g t lim t a g t and lim t a g t If any of the quantities do not exist explain why Solution A limit does not exist when the corresponding right hand and left hand limits are not equal 7 Sketch the graph of the function and use it to determine the values of a for which limx a f x exists a 4 1 2 4 g a 3 4 1 Not defined limt a g t 3 4 1 2 limt a g t 2 4 5 2 limt a g t Does not exist 4 Does not exist 2 3 x x 2 x2 2 2 x 3 10 x x 3 f x Solution From the picture we see that the second and third pieces of the function are glued together seamlessly at x 3 so limx 3 f x exists and is equal to 7 The limit of f x at x 2 does not exist since there is a jump discontinuity at x 2 The function f x is continuous at all other values of x 4 8 Consider the function f x tan cid 0 1 cid 1 x a Show that f x 0 for x 1 1 2 1 3 Solution Plugging in these values of x we have f 1 tan 0 f 1 2 tan 2 0 and so on b Show that f x 1 for x 4 4 5 4 9 Solution Likewise f 4 tan 1 4 tan 4 1 and so on c What can you conclude about lim x 0 tan cid 18 1 cid 19 x Solution We have found two sequences of numbers approaching x 0 from the right where f x approaches two different values 0 and 1 This means that limx 0 tan 1 x does not exist 5
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