# 4.0 5-Minute Review: Rational Functions

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1 mth 130 dy 4: working with limits Minute Review: Rtionl Functions DEFINITION A rtionl function 1 is function of the form y = r(x) = p(x) q(x), 1 Here the term rtionl mens rtio s in the rtio of two polynomils where p(x) nd q(x) re polynomils The domin of rtionl function consists of ll vlues of x such tht q(x) = 0 EXAMPLE 41 Here re severl exmples () r(x) = 2x2 + 3x + 1 4x x 2 (b) s(x) = 1 2x + 7 (d) p(x) = 6x (e) r(t) = 5t + t 2 t(x) = 3x + 1 x x 4x + 10 And some non-exmples (why ren t these rtionl functions?): () r(x) = sin(2x2 + 1) 3x 4x x 2 (b) s(x) = 2x + 7 t(x) = 3x1/2 + 1 x Cution: Often times we will need to simplify expressions involving rtionl functions Be creful Do you see why r(x) = x2 4 (x 2)(x + 2) x 2 = 2x x(x 2) nd s(x) = x + 2 x re not the sme function? The domin of r(x) is ll x = 0, 2 while the domin of s(x) is ll x = 0 The two functions hve the sme vlues t ll points where both re defined YOU TRY IT 41 Solve the following () Determine ll x for which x2 4x 5 x 2 1 (b) Do the sme for x3 3x 2 + 2x x 2 x nd x 2 nd x 5 hve the sme vlues x 1 Crefully sketch the grph of x3 3x 2 + 2x x 2 Think bout prt (b) x 41 5-Minute Review: Conjugtes Conjugtes re usully discussed in reference to expressions involving squre roots nd typiclly hve the form + b nd b, where cn be ny expression For exmple, x + 3 nd x 3 Conjugtes re useful becuse when you multiply them together, the middle terms cncel: eg, ( + b)( b) = 2 b or ( x + 3)( x 3) = x 3 Simplify ech of these expressions by multiplying both the numertor nd denomintor by n pproprite conjugte () x 5 x 5 (b) 4 x + 2 x 2x 18 x 3 (d) x + h x h

2 mth 130 dy 4: working with limits 2 SOLUTION For prt (b) use 4 x x = 4( x x) = 4( x x) = 2( x x) x + 2 x x x x + 2 x 2

3 Clculting Limits 42 Bsic Limit Properties There re severl bsic limit properties which ese the problem of clculting limits Most of these properties re quite strightforwrd nd would be wht you might suspect is true In more dvnced clculus courses (clled nlysis courses) ech of these properties would be crefully proven The first property tht we list is lmost tutology It concerns the liner function f (x) = x THEOREM 41 lim x = The limit is sking: As x pproches, wht is y = f (x) = x pproching? Well, obviously! This is lso geometriclly cler in the left-hnd pnel of Figure 41 b Figure 41: Left: As x pproches, x pproches so lim x = Right: For As x pproches, the constnt function f (x) = b is lwys equl to b so lim b = b THEOREM 42 If b is ny constnt, then lim b = b The limit is sking: As x pproches, wht is y = f (x) = b pproching? Well, since b is constnt, not is f (x) pproching b, it is ctully equl to b This is geometriclly cler in the right-hnd pnel of Figure 41 The limit is sking: As x pproches, wht is y = f (x) = b pproching? Well, since b is constnt, not is f (x) pproching b, it is ctully equl to b This is geometriclly cler in the right-hnd pnel of Figure 41 We cn combine these two theorems in more generl result bout lines or liner functions THEOREM 43 (Liner functions) Let, b, nd m be ny constnts nd let f be the liner function f (x) = mx + b Then lim f (x) = lim mx + b = m + b = f () Think bout wht this sys: s x, mx m, so mx + b m + b We hve ctully used two mthemticl opertions, multipliction nd ddition, nd sid tht the limit opertion intercts with them in specil wy: the order of Remember from Lb 1, tht function where lim f (x) = f () is clled continuous function t

4 mth 130 dy 4: working with limits 4 opertions does not mtter The results in the next section this cler in number of situtions EXAMPLE 42 Determine the following limits () lim x 4 f (x) where f (x) = 2x (b) lim g(t) where g(t) = 7 t 2 2 t 8 lim h(x) where h(x) = 9 x 1 ) SOLUTION () lim f (x) = lim ( 2x + 1 Liner x 4 x 4 2 = f (4) = 15 2 ( ) (b) lim g(t) = lim 72 Liner t 8 = g( 2) = 15 t 2 t 2 lim h(x) = lim 9 Liner = 9 (Here m = 0 or we could use the constnt theorem) x 1 x 1 Properties Involving the Order of Opertions The next severl properties involve the interchnge of the limit opertion with other rithmetic opertions such s ddition or multipliction These properties re importnt becuse they gretly simplify the clcultion of limits It is importnt to recognize tht not ll mthemticl opertions cn be interchnged in this wy For exmple, the squre root of sum is not the sum of the squre roots: + b = + b In prticulr, = 5 while = 7 Clerly, the order of opertions mtters; tking the squre root before or fter summing chnges the result However, the limit properties below indicte tht certin mthemticl opertions with limits will give the sme nswer, regrdless of the order in which they re crried out This is specil nd useful! Note the order of opertions in ech prt THEOREM 44 (Order of Opertions) Assume tht lim f (x) = L nd lim g(x) = M nd tht b is constnt Then 1 (Constnt Multiple) lim b f (x) = b(lim f (x)) = bl 2 (Sum or Difference) lim( f (x) ± g(x)) = lim f (x) ± lim g(x) = L ± M The limit of sum is the sum of the limits 3 (Product) lim f (x)g(x) = (lim f (x)) (lim g(x)) = LM The limit of product is the product of the limits f (x) 4 (Quotient) lim g(x) = lim f (x) lim g(x) = L, s long s M = 0 The limit of quotient is M the quotient of the limits EXAMPLE 43 Determine the following limits Indicte which limit properties were used t ech step Assume tht lim f (x) = 4 nd lim g(x) = 3 Evlute x 2 x 2 () lim[2 f (x) + 3g(x)] Sum = lim 2 f (x) + lim 3g(x) x 2 x 2 x 2 2(4) + 3( 3) = 1 (b) lim 2 f (x)g(x) Prod = lim 2 f (x) lim g(x) x 2 x 2 x 2 2(4) ( 3) = 24 = 2 lim x 2 f (x) + 3 lim x 2 g(x) = = 2 lim x 2 f (x) lim x 2 g(x) = 6x + 7 f (x) Diff, Quot lim lim = x 2 (6x + 7) lim x 2 f (x) Liner = 19 4 = 5 x 2 g(x) lim x 2 g(x) 3

5 mth 130 dy 4: working with limits 5 (d) lim x 2 = lim x x Prod = lim x lim x Thm = 1 = 2 More generlly, if n is positive integer, then lim x n = n by using the product limit lw Tht mens the the function f (x) = x n is continuous for every point THEOREM 45 (Order of Opertions, Continued) Assume tht lim f (x) = L nd tht m nd n re positive integers Then 5 (Power) lim [ f (x)] n = [lim f (x)] n = L n 6 (Frctionl Power) Assume tht m n is reduced Then [ ] n/m lim [ f (x)]n/m = lim f (x) = L n/m, provided tht f (x) 0 for x ner if m is even EXAMPLE 44 Determine lim x 3 (4x 1) 5 Indicte which limit properties were used t ech step SOLUTION lim (4x x 3 10)5 Powers = [lim 4x 10] 5 Diff = [lim 4x lim 10] 5 x 3 x 3 x 3 = [4 lim x 3 x lim x 3 10] 5 Thm 51,52 = [4(3) 10] 5 = (2) 5 = 32 EXAMPLE 45 Determine lim x 2 1 Indicte which limit properties were used t x 2 ech step SOLUTION Notice tht x 2 1 = (x 2 1) 1/2 is frctionl power function In n the lnguge of Theorem 45, m = 1 2 is reduced nd m = 2 is even Ner x = 2, f (x) = x 2 1 is positive So Theorem 45 pplies nd we my clculte the limit s ( ) 1/2 lim x 2 Frc Pow Poly 1 = lim x 2 x 2 x2 1 = (3) 1/2 = 3 EXAMPLE 46 Determine lim x 3 (2x2 x + 1) 4/3 Indicte which limit properties were used t ech step SOLUTION (2x 2 x + 1) 4/3 is frctionl power function with m n = 4 3 which is reduced nd m = 3 is odd Ner x = 2, f (x) = 2x 2 x + 1 is positive So Theorem 45 pplies nd we my clculte the limit s ( ) 1/2 lim 2x 2 Frc Pow Sum, Prod x + 1 = lim x 3 x 3 2x2 x + 1 = ( ) 1/2 = 4 EXAMPLE 47 Determine lim x 3 (x2 25) 3/4 SOLUTION (x 2 25) 3/4 is frctionl power function In the lnguge of Theorem 45, n m = 3 4 is reduced nd m = 4 is even Ner x = 3, x2 25 is negtive Since (x 2 25) 3/4 is not even defined ner 3, this limit does not exist

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