2.2 - Infinite Limits

2.2 - Infinite Limits s Limits C Intinity (1) Infinite Limits When x a, but f(x) 100 i.e. lim (f(x)) = I 8 [ VERTICAL ASYMPTOTES (2) Limit at infinity when X IOO, and f(x) ??? it depends on function! but here we may be talking about horizontal asymptotes. we can't divide by o, but what happens if we divide by numbers that are close to o? Consider the function F(x) = and the following table: X x (-) 0.1 (-) 10 (-) 0.01 (-) 100 (-) 0.001 (-) 1000 we say that lim (-) but we could also 10L ...O from the LEFT, I.C in that case Lim 40 (

we also notice on the graph at y= = x there is a vertical asymptote at X=O ex : Consider the following Limit: x-3 x-3 lim (xt) Since we can't divide by O, we can't let x = 3 X-3-0 So.. Instead let's investige the LEFT RIGHT Limits RIGHT I'm x*3+ ( 4 ) 4 X-3 o 6 x- 3+ = x 3 3 = x-3>0 = x-3 - 0+ LEFT lim (is) 4 = -00 x-3 ( 0- x 23 = x 3 3 X-3-5 x-5 o+ = (x+5)2 0+

LEFT x*.5 I'M ( (x+5)' x+7 ) 4 X- - 5 If x - 5 X+540 x+5 o- (x.5)3 o+ ex:(6+5)2 = 1 : x -5 lim (x.).) = 8 (x+5)2 The graph of y x+7 has a V.A at X--5 (x.5)2 -5 Definition If f(x) I 00 as x at or x a then the line x = a is a vertical asymptote These typically occur whenever fax KFO as x-a O ex: where are the vertical asymptotes of the function g(00) = x2+2x-8 ? x2 6x - 8 Sol'n where IS the denominator equal to o? x2-6x-8=0 (x. +)(++2) : 0 X+9=0 OR x+2 x=-2 @ x= -4 x-a lim o indeterminate = common factor x2+bx+8 O (x+4)(x-2) (x+4)(x-2)

= X>-4 I'm ( (x-A)x-23 (x4 a)(x.2) ) = x-4 lim = -4-2 -4+2 = -6 -2 = =13 No vertical asymptote @ x = - 4 @ x = -2 lim (***2x-8 ) - 8 UNDEFINED X-2 0 LEFT X+-2 lim (xy) (x-2) (x12) = x - 2 lim ( x-2) x+2 -2-2 o = + 00 6 x- - 2 o- g(x) 8 as x- -2- There is a V.A at x=-2 = Limits At Infinity These are easier / more obvious: lim ) 1 Some important limits at infinity x s x for n= 1,2,3,4, ( pos. whole #) lim (x") = + 8 lim (xn) = X-00 { T 8 if n is EVEN - 00 if n IS ODD

lim (2) = 0 The limit will be equal to zero X-30 These make it easier to evaluate limits at infinity of polynomials: ex lim ( 2x3 + 3x2 - why do these equal 0? X-00 = lim (2x3 ( 2x3 2x3 2nd 3x2 2x3 ) X->00 because: lim 3/2 2x 3 1 = lim (2x3 (1 7 2x 2x2 2x3 D) o = x lim 8 ( 2 3 x ) = lim (2x3) = 3 X300 2 x->00 lim (1) = 00 =3.0 = 2 = 0 We extend our knowlidge of limits @ infinity of paynomias to limits @ infinity of RATIONAL FUNCTIONS ex: iim ( 2x2 3x-10) bx3 -10x+18 Since the numerator & denominators are both poynomiais we will factor out the LARGEST POWER of x from each poynomial 3x 10 lim 8 (2x2 (201 10 2x ) = lim bx3 ( bx - 9, ) (2x2 (1- / 6x3 (1- ) = lim - 00 (2x) bx3 = lim (1) = 1 lim 3x 3 X-200 (L) = 0 =01 =

ex: x-o- lim ( -x- ii) -5(x3)2 = I'm ( 5x2 - x X-00 -5(x*+6x+9) ) = lim ( -5x2-30x-45 5x2-, ) = I'm ( -5x2 5x2 ( (---) so. ,) = lim 5x2 5x2 (1+%) (1-5%) ) = lim ( -5x2 5x2 ) = lim (-1) = -1 // Definition If lim f(x) = L a finite real number then there is a line is a horizontal asymprate If lim f(x) = I 8 then there is NO horizontal asymptote A final "bonus' limit @ infinity f(x) = ( + 1)* = (**) x we could show that x+00 lim ( x + ) = 1 (x(1-)=(=) x = = 1 Proof So what happens if we then apply ( ) to that value that is "close to" 1? This defines the number we call " C" e is like n & 52, X - 00 lim (1+ + 1/4 x )* = e si 2.71828 it's irrational!