2.6 Limits at Infinity, Horizontal Asymptotes Math 1271, TA: Amy DeCelles. 1. Overview. 2. Examples. Outline: 1. Definition of limits at infinity

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1 .6 Limis a Infiniy, Horizonal Asympoes Mah 7, TA: Amy DeCelles. Overview Ouline:. Definiion of is a infiniy. Definiion of horizonal asympoe 3. Theorem abou raional powers of. Infinie is a infiniy This secion is abou he long erm behavior of funcions, i.e. wha happens as ges really big posiive or negaive. Someimes he funcion will approach a specific number as ges big. If a funcion f approaches a specific number L as ges larger and larger posiive, we say ha he i of f as approaches infiniy is L and we wrie: f L Similarly, if f approaches L as becomes larger and larger negaive, f L In he firs case, he graph of f flaens ou on he righ, and i approaches he line y L. In he second case, he graph flaens ou on he lef, approaching he line y L. In eiher case his line is called a horizonal asympoe of f. The following heorem is helpful in evaluaing infinie is: Theorem: If r is a posiive raional number, r 0 Addiionally, as long as r is defined for negaive values of, r 0. Eamples Eample : Evaluae he following i: 3 +

2 Noice ha as ges large, he op ges large and he boom ges large, so we can immediaely ell wha he behavior of he funcion is as ges large. We can hink of his as a compeiion beween he numeraor and he denominaor. Boh are geing larger and larger, and he quesion is: Does one of hem ousrip he oher? Or do boh he numeraor and he denominaor say abou even as hey ge larger? In order o figure his ou, we need o use he following echnique: divide he numeraor and denominaor by he highes power of in he denominaor. So here, we will divide op and boom by : Now we can evaluae he i of he op and he i of he boom: So we can conclude ha he numeraor and he denominaor sayed abou even as hey grew larger. This is because he highes power of in he numeraor is he same as he highes power of in he denominaor. Noice ha his also means ha he funcion f 3+ y 3. Eample : Evaluae he following i: has a horizonal asympoe: Noice ha he i of he op is infinie and he i of he boom is also infinie. So we mus divide he numeraor and denominaor by he highes power of in he denominaor, which in his case is 3 : Since he i is zero, we can conclude ha he denominaor ousripped he numeraor. This is because he highes power of in he numeraor is smaller han he highes power of in he denominaor. Noice ha his means ha he funcion f + y has a horizonal asympoe

3 Eample 3: Evaluae he following i: Since he i of he op is infinie and he i of he boom is infinie, we mus divide he numeraor and denominaor by he highes power of in he denominaor, which in his case is : So in his case he numeraor ousripped he denominaor. This is because he highes power of in he numeraor is greaer han he highes power of in he denominaor. Eample : Evaluae he following i: This funcion is differen from he hree previous eamples because i is no a raional funcion. So we have o be a lile more careful. The fac ha here is a square roo in he denominaor makes hings more complicaed. We have o ake he square roo ino accoun when we re looking for he wha power of we should divide by. Looking a he denominaor by iself we can ask, Which erm is mos imporan, when ges large? In his case, he 9 is mos imporan, because i will keep geing larger and larger, evenually overwhelming he conribuion of he +. So if we ignore he +, we can say ha when is large, he denominaor is approimaely 9, i.e. i is approimaely 3. So we ac as if is he highes power of in he demominaor. So we divide op and boom by and wach our algebra carefully!

4 Since is posiive,, and we can do he following: Eample : Find he horizonal and verical asympoes of f. To deermine wheher here are horizonal asympoes we mus evaluae he is a infiniy. f This means ha he line y is a horizonal asympoe. Similarly, we can compue ha: f So y is he only horizonal asympoe. To deermine wheher here are verical asympoes we check o see where he denominaor is zero. If we facor he denominaor, we see ha: + so he denominaor is zero when and. So and are possible verical asympoes. If we facor he numeraor as well, we can see: f So here will be a hole a, because:

5 Bu here will be an asympoe a because: + + and as ges close o from he lef he numeraor ges close o posiive and he denominaor ges close o zero and is negaive. So, Eample 6: Find he horizonal and verical asympoes for f +e e. To find he horizonal asympoes we look a he is a infiniy. We know ha e ges larger and larger as, so: + e e We know ha e ges close o zero as, so: e + e e e So he lines y and y are horizonal asympoes. To find he verical asympoe we noice ha he denominaor is zero when 0, and, so when 0, he denominaor is posiive, close o zero and he numeraor is posiive, close o 3, so: + e 0 e and he line 0 is a verical asympoe.

CHARGE AND DISCHARGE OF A CAPACITOR

CHARGE AND DISCHARGE OF A CAPACITOR REFERENCES RC Circuis: Elecrical Insrumens: Mos Inroducory Physics exs (e.g. A. Halliday and Resnick, Physics ; M. Sernheim and J. Kane, General Physics.) This Laboraory Manual: Commonly Used Insrumens: