2.1 - Introduction to Limits

2.1 - Introduction to Limits the value of a fractionat x = a, the Output of the function when x=a f(a) "f of a on the graph of a function f(a) is the y coordinace at xa 三 NS the limit of a function as X APPROACHES (what do the y coordinates. approach) a. xa (with x + a) is the value that a approach ES, Notation (aka the symDOIS) let f(x) be a function we write : lim(f(x)) = L xa to mean the limit of f(x) as x approaches a is L" OR f(x) Las x a If the value of f(x) gers closer and closer to Las x gets closer and closer to a (with x ≠ a) ex: Desmos Graph (1) lim (x²) = X2 4 lim (f(x)) = ? X2 It depends on How We approach 2 DEFINITION: ONE-SIDED LIMITS LEFT HAND LIMIT X approaches a FROM the left lim (f()) Xa- RIGHT HAND LIMIT X approaches a FROM the right lim (f(x)) xa+ Note "a itself may be positive or negative we may see things like: lim (f(x)) Jim (f) x-3 Theorem: lim (f(x)) = L xa if and only if lim (f(x)) xa = lim (f(x)) = xa L this means that: lum (f(x)) ≠ lim (f(x)) xat then the lim (EOS) DNE xa ex from desmos lim (foo) = 3 X2 Since 377 lim (f(x)) DNE X2 lim (f(x)) = 7 x> 2+ ex: lim/x1 = we can see this graphically lim x1 X IXI follows a different rule depending on how we approach O we must 100k at left limits Fright 100king at left limit: x → 0- if SO 1x1= -x lim = lim () -= lim (-1) = -1 if x 20m() 100king at the right limit: 0 lim 1x1 X = lim x= X lim (1) = 1 30 1x1= x DNE ex: lim 1x1 = O we can see this graphically lim x1 (1) IXI follows a different rule depending on how we approach O > we must look at left lright. limits looking at left limit: x → 0- if SO 1x1= -x -x = lim (-1) = -1 -OX 100king at the right limit: x→0 + if XW0 30 1x1=X lim JDNE lim 1x1 X lim x= XO lim (1) = 1 Properties of Limits Let f & g be two functions such that lim (f(x)) = L xa lim (g(x)) = M (1) lim (k) = K xa limit of a constant is the constant (2) lim (x) = a xa what does x approach as x approaches a?" (3) lim (f(x) = gu)) = lim (f(x)) I lim (g(x)) = L+ M lim (12-x) = lim (12) lim (x) = x77 12-75 tex (4) lim (K-f(x)) = K lim (f(x)) lim (6x-1) = lim (6x) - lim(i) = 6. lim (x) lim() メニ = 6(8)-1 47 8 84x = (5) lim (f(x)+g(a)) = lim (f(x)) lim (gas) = lim (x²) = a² lim (x²) = lim (xx) = x2 lim (x) lim(x) = 2.2 = 2 x2 Tex Cx : polynomial lim x3+3x-4 Paynomial num: lim (x3+3x-4) = (1)33(1)-4 = 0 denom: lim (x² + 1) = (1)²+1=20 lim X x+3x-4 لام ۱۰ 2 Theorem Since rational functions are ratios (or quotients) of polynomials, we also have that: lim (r(x)) = lim P(x) 9(x) = P(a) ৭(α) only il qua) ≠ 0 ex lim (x²-x-6) = x-2 = lim (x²) lim (x) - lim (6) x-2 2 -2-(2)-6= 4+2-60 lim x²+1 (x+1)2 لا ( num lim (x²+1) = (1) 21 = 2 denom. lim (x+1)² = (1+1)² = 470 2 lim x²+1 (x+1)²/ = lim x²+1 (x+1)2 lim x-3 k+3 x²+ 3x num lim (x+3)=-3+3=0 X-3 denom lim (x²+3x(=)-3(+)-3( = ۹-۹ - ه x-3 Lim x-3 x+3 x3x = we call this an INDETERMINATE FORM. Definition If lim (f(x)) lim (g(x)) = 0 xa then lim/f(x) 9(x) is a indefinite form" Theorem If f(x) is a polynomial and if f(a) then (x-a) is a FACTOR of f(x) Back to lim -3 x+3 x²+3x = lim 3 x+3 x(x+3) (x+3) x(x+3) as long as x=-3 because (-3+3) = ex lim 25 ) x²+3x-40 num: lim x5 (x²-2 25)(5)²-25 = 0 denom: lim (x²+3x-40) = (5)²+3(5)-40-0 X5 we have a O indeterminate form So... let's factor the common factor of: x-5 lim x5 x²-25 x²+3x-40 lim x5 (x+5) (x+8) 10 13 lim x+5 X+8 = lim (x+5)= lim (x+8) 5+5 5+8 * NEVER SKIP THESE STEPS TO FIND INDETERMINATE FORM The Difference Quotient of a function. For any function f(x), the difference quorient is the expression: f(xh) - f(x) ox. Find the difference quotient for the functions: f(x) = x²-2x f(x)=(xh)² - 2(x+h) = (x+h)(x+h) -2x+n = x²+2xh+h²-2x-2n • f(x+h) - f(x) = x²+2x+h²-2x-2-(x²-2x) = x²+2xh+h²-2x-2n-x²+2x 2xh+h2h •f(x+h)-f(x) = 2xhh²2h (2) f(x) = 2x3 +1 (3) f(x) = 2-2