Limits
From last time... Let y = f (t) be a function that gives the position at time t of an object moving along the y-ais. Then Ave vel[t, t 2 ] = f (t 2) f (t ) t 2 t f (t + h) f (t) Velocity(t) =. h!0 h f(a+h) f(a) h a We need to be able to take its! a+h
From last time... Let y = f (t) be a function that gives the position at time t of an object moving along the y-ais. Then Ave vel[t, t 2 ] = f (t 2) f (t ) t 2 t f (t + h) f (t) Velocity(t) =. h!0 h f(a) f(a+h) h a We need to be able to take its! a+h
From last time... Let y = f (t) be a function that gives the position at time t of an object moving along the y-ais. Then Ave vel[t, t 2 ] = f (t 2) f (t ) t 2 t f (t + h) f (t) Velocity(t) =. h!0 h f(a) f(a+h) h a a+h We need to be able to take its!
Limit of a Function Definition We say that a function f approaches the it L as approaches a, written f () =L,!a if we can make f () as close to L as we want by taking su ciently close to a. f() L Δy Δ i.e. If you need y to be smaller, you only need to make smaller ( means change ) a
One-sided its f(a) L r L l Right-handed it: L r =!a + f () if f () gets closer to L r as gets closer to a from the right a Left-handed it: L` =!a f () if f () gets closer to L` as gets closer to a from the left Theorem The it of f as! a eists if and only if both the right-hand and left-hand its eist and have the same value, i.e. f () =L if and only if!a!a f () =Land f () =L.!a +
Eamples 2!2 +3 =0 2! =2!0 is undefined 8 8 8 4 4 4-4 4-4 -4 4-4 -4 4-4 -8-8 -8
Theorem If!a f () =Aand!a g() =B both eist, then.!a (f ()+g()) =!a f ()+!a g() =A + B 2.!a (f () g()) =!a f ()!a g() =A B 3.!a (f ()g()) =!a f ()!a g() =A B 4.!a (f ()/g()) =!a f ()/!a g() =A/B (B 6= 0). In short: to take a it Step : Can you just plug in? If so, do it. Step 2: If not, is there some sort of algebraic manipulation (like cancellation) that can be done to fi the problem? If so, do it. Then plug in. Step 3: Learn some special it to fi common problems. (Later) If in doubt, graph it!
Eamples 2.!2 +3 2 2.! 2 = 0 because if f () = +3,thenf(2) = 0.!0!0 If f () = 2,thenf() isundefinedat =. However, so long as 6=, So 3.!0 f () = 2! p +2 p 2 = ( + )( ) = +. 2 = +=+= 2.!!0, so again, f () isundefinedata.
Eamples 3.!0 p +2 p 2!0, so again, f () isundefinedata. Multiply top and bottom by the conjugate:!0 p +2 p 2 =!0 p +2 p 2 +2 2 =!0 ( p +2+ p 2) =!0 ( p +2+ p 2) =p p =!0 +2+ 2 2 p 2! p p! +2+ 2 p p +2+ 2 since (a b)(a + b) =a 2 b 2
Eamples.! 2 3+2 2 +4 5 = 6 2 2.! 2 = -2-2 3.!0 is undefined - -2 4.!0 (3 + ) 2 3 2 = 6
Infinite its If f () gets arbitrarily large as! a, thenitdoesn thaveait. Sometimes, though, it s more useful to give more information. Eample: For both f () = and f () =, 2!0 f () does not eist. However, they re both better behaved than that might imply: =,!0 +!0!0 =!0 + does not eist!0 Why? A vertical asymptote occurs where f () =± and!a +!a =, 2!0 2 = f () =± 2 =
Infinite its Badly behaved eample: f () =csc(/) -0.5 0.5 Zoom way in: -0.005 0.005 (denser and denser vertical asymptotes) csc(/) does not eist, and!0 +!0 csc(/) does not eist
Limits at Infinity We say that a function f approaches the it L as gets bigger and bigger (in the positive or negative direction), written f () =L or f () =L!! if we can make f () as close to L as we want by taking su ciently large. (By large, we mean large in magnitude) Eample :! =0 and! =0.
Limits at Infinity: functions and their inverses Theorem If!a ± f () =, then! f () =a. If!a ± f () =, then! f () =a. Eample: Let arctan() be the inverse function to tan(): y =tan(): -3π/2 -π/2 π/2 3π/2
Limits at Infinity: functions and their inverses Theorem If!a ± f () =, then! f () =a. If!a ± f () =, then! f () =a. Eample: Let arctan() be the inverse function to tan(): y =tan(): -π/2 π/2 (restrict the domain to ( 2, 2 )sothatwecaninvert) y =arctan(): π/2 -π/2
Limits at Infinity: functions and their inverses Theorem If!a ± f () =, then! f () =a. If!a ± f () =, then! f () =a. Eample: Let arctan() be the inverse function to tan(): y =tan(): -π/2 (restrict the domain to ( π/2 2, 2 )sothatwecaninvert) Since! /2 = and! /2 + = y =arctan(): π/2 -π/2 we have! = /2 and! = /2
Rational functions Limits that look like they re going to can actually be doing lots of di erent things. To fi this, divide the top and bottom by the highest power in the denominator! E E 2 E 3!! 3 +2! Fi: multiply the epression by /3 : / 3! 3 +2 =! 3 +2 / 3 / 3 3 2 2 +!! 4 2! Fi: multiply the epression by /2 / 2 4 2 +2!! 3 +3! Fi: multiply the epression by /3 / 3 =! 2 3 + 2 = 0 0 +0 = 0 3
Rational functions: quick trick! 3 +2 =0 3 2 2 +! 4 2 = 3 4 4 2 +2! 3 = +3 Suppose P() =a n n + +a +a 0 and Q() =b m m + +b +b 0 are polynomials of degree n and m respectively. Then in general, 8 P()! Q() = a n n! b m m = a >< 0 n < m n n m = a n! b b m >: m n = m ± n < m
Eamples: Other ratios with powers. Eample: p! 3 2 +2 [hint: multiply by / / and remember ap b = p a 2 b.]! p 3 2 +2 =! =! =! =! = p 3 2 +2 p 3 2 +2 q (3 2 + 2) 2 q 3+ 2 2 p = p 3+0 3 / /