| Start | so last time we did limits as x approaches infinity and you want to make sure youre getting good at these so were gonna review just a little bit more so the standard idea, whats going on with infinity, theres two different things one is what makes you go to infinity, also known as vertical behavior what happens when you go out of infinity which is end behavior, another words |
| 0:44 | say you have a graph that looks something like that those are 4s if you cant read my hand writing youre allowed to make fun of my handwriting |
| 1:04 | so were looking for a limit as approaches -4 minus sign remember the minus side is this side of -4 so you get a curve you go towards where x is negative 4 and you keep going up up up continues going up it has an asymptote so f would equal positive infinity |
| 1:35 | thats what we mean when we say x is minus 4 minus when you say x is -4 plus we know mean when your coming from this side of the -4 so numbers are just bigger then negative 4 but only a little bit so if you were thinking of plugging you would say im plugging in negative 4 plus a tiny amount or negative 4 minus a tiny amount if your on the other side so at negative 4 and going from this direction and you go down the curve |
| 2:05 | so the limit as x approaches -4 ffrom the plus side, thats f of x equals minus infinity are those the same thing? no they are not because one is plus infinity one is negative infinity so the limit as x approaches -4 |
| 2:30 | of f of x does not exist and i know on some of the webassigns it wanted you to put none and instead wanted to put dne if you read the problem careflly it would tell you which one to put i know thats very frustrating dont know what to tell you alright well now whats going on if we go to positive 4 |
| 3:00 | well when we approach 4 from the minus side this side of 4, were going up to positive infinity so its positive infinity and what do we do if we do x approaches 4 from the other side well now were going down to minus infinity so were going down on the curve |
| 3:37 | where you end up here you just keep going down and down deeper and deeper in the hole of darkest, scared because its sophmore year and then you get minus infinity so these two dont agree so once again |
| 4:06 | so if we just ask for the limit as x approaches 4, we didnt give you a side it would be does not exist so thats, when you create a number you get out negative inifinty remember inifinty isnt a number, another words what you get out is the number keeps ggrowing |
| 4:30 | it just keeps going up forever or down forever what happens if i want to know what happens after infinity well now if i do the limit x apporaches infinity im saying what haappens when im on the curve and i keep going out, bigger and bigger numbers of x |
| 5:08 | and of course at minus infinity im also going to approach 0 notice their is no sign to an infinity limit like i said last week on monday there isnt a plus or minus side |
| 5:30 | to infinity your either going to infinity or going to negative infinity because if theres a side you go to infinity and come back from another direction cant happen these two could agree cause youd get a curve that looks something like this now when you get plus 4 from the minus side you get plus infinity |
| 6:03 | and when you approach four from the right side you get plus infinity so the limit as x approaches 4 would be infinity again so you ge the limit as x approaches 4 or the minus side of f of x youd get pi/infinity and youd get the limit as x approaches 4 from the plus side of x you would also get positive infinity therefore |
| 6:32 | the limit as x approaches 4 is infiinty
so far so good?
|
| 7:00 | now lets do some algebraic/numeric types so people were getting this a little confused at the end of class the other day so lets go through this again |
| 7:38 | suppose you had that limit, limit as x goes to infinity 5x cubed minus 7x plus 1 on top 2x cubed plus 4x squared plus 1 on the bottom the numerator and the denominator |
| 8:00 | so does everyone know what that is? we did the rule the other day its 5/2 now if you see this on the exam you can not just write 5/2 however you dont have to do a lot of work what you would need to do, first you would write as x approaches infinity the terms |
| 8:48 | you write something as x approaches infinity the terms other than 5x cubed and 2x cubed are insignificant, so we can ignore them so this limit |
| 9:01 | behaves like this if you write that youd get perfect credit, another words you dont have to divide by x cubed you could do all that stuff we did last time why write something like that because we dont know if you just wrote down 5/2 and youre not memorizing the rule and you have no clue whats going on or copy off the person neck to you and wrote 5/2 and wrote it larger |
| 9:32 | when you are doing your exam you should remember you are giving your answers youre explaining yourself to the person on the other side going to be a ta whos going to grade 100's of these and is not going to be in the mood by the time he gets to ypurs for trying to figure out what the heck you are up to so you want to make sure its clear so that person could be me by the way, ill be grading one question so if im tired and im on paper 500 |
| 10:04 | i dont want you to try and get away with that so demonstrate your knowledge somehow you can demostrate your knowledge very simple but you essectially want to say, is when youre doing these kind of ratios the only thing that matters is the highest term on top, the highest power term on top in the numerator and the highest term in the denominator you can say top you can say bottom |
| 10:32 | you can say treble and base whatever you want however we dont want to confuse this the limit as x is not going to infinity |
| 11:25 | this is a totally different type of rational expression |
| 11:34 | that works, lets see what happens, i just made this up i might make a mess of things but we will find out what do you do when you get something like this yea i like hearing thtat from the front so 25 plus 11 is 36 the square root of 36 is 6 is 0, so you plug is 5-5 and get 0 and then you cry |
| 12:03 | you say what do i do? and rationalize how do i rationalize? i multiply the top and the bottom by the conjugate of that the conjugate of the square root x squared plus 11, minus 6 is the suqrae root of x squared plus 11 plus 6 |
| 12:30 | the idea of the conjugate is when you take the one binomial, a plus b and you multiply it by the conjugate a-b you get a suqared minus b suared so in this case, im gonna erase this, thats the idea of the conjugate im gonna erase and i erase alright so you got to get the conjgate so you ultiply this by this |
| 13:03 | you get x squared plus 11 minus 36 and on the bottom dont try and combine terms just leave it alone doesnt matter |
| 13:41 | so now x squared plus 11 minus 36 simplifies as x squared minus 25 youre still ignoring whats on the bottom notice if you still plug in 5 you still get 0 on top and 0 on the bottom |
| 14:04 | so your life isnt perfect yet we can now factor the top |
| 14:46 | alright now we cancel the x minus 5 and we get the limit as x approaches 5 whats left over, now you could now youre not gonna have a 0/0 problem, when you plug in 5 on top you get 10 doesnt matter you no longer have a 0/0 problem |
| 15:01 | and whne you plug in 5 on the bottom
25 plus 11 is 36, 36 plus 6
is 12
okay
you have homeowrks like webassign that looks like this
i got that right, any mistakes?
|
| 16:06 | alright suppose we wanted to do this one so whats the limit as x approaches infinity of 1 plus e to the 1/x well as x approaches infinity 1/x approaches 0 |
| 16:34 | e to the 0 1 so as x approaches infinity this becomes 0, e to the 0 is 1 1+1 is 2 and what about when x approaches minus infinity its going to be the same thing okay |
| 17:00 | easty ones again everybody see where thats ccoming from when x is infinity 1/infinty the limit of x approaches infinity is 1/x is 0 the limit of x approaches infintiy is 1/x |
| 17:32 | thats a 0 thats because the bottom gets bigger and bigger so the whole thing get smaller as x approaches infinity 1/x appproaches 0 so that means this is becoming e to the 0 e to the 0 is because anything to the 0 is 1 so 1+1 is 2 |
| 18:00 | so far so good? what about when x approaches 0 well what is the limit as x approaches 0 from the plus side of 1/x well now the bottom is getting smaller and smaller so 1/x is getting larger and larger its staying the positive side of 0, so 1/x is positive so this is going to approach positive infinity |
| 18:36 | and 1 plus infinity is infinity what about if you approach 0 from the minus side well now we get minus infinity why are we gonna get minus infinity |
| 19:02 | well as we get closer and closer to 0, 1/x gets bigger and bigger
but were using negative numbers
so were gonna get closer to negative
what is e to the negative infinity
anybody know?
e to the -1 is e to the plus 1/x so as we approach positive infinity we are getting closer to 1 over e to the infinity so thats gonna approach 0 this is gonna approach 0 so the whole thing is gonna go to 1 |
| 19:34 | so if were doing this term 1/x as x goes towards 0 from the minus side this term will go close to negative infinty e to the negative infinty really means 1/ e to the positive infinity |
| 20:01 | thats just 1 over a really big number so that just a really small number so this will get closer and closer to 0 the e term so the whole thing will approach 1 still see some glazed eyes out there |
| 20:36 | everything we do is in radians all the trigonometry is in radians wen you cant get your trigonmetry stuff to work right first thing you do is check your calculator and see if its in degree mode put it in radian mode if your using some app on your phone figure out how to put it in radian mode cause if you email me ill tell you to check and see if you did it in degrees theres a webassign problem thats the tangent of something over the tangent of something |
| 21:04 | your doing it in degrees you getting the wrong answer
how do we feel about the whole limit to infinity thing?
|
| 21:34 | you guys dont get what hes asking, ill repeat this
the best thing to understand whats going on with 1/x is look at the graph of 1/x
the graph of 1/x
looks like this
this is the graph, y=1/x
so what happens when x is a really big number?
when you have a really big number you put it in the denominator of a faction and its smaller |
| 22:03 | 1/100. 1/1000, 1/a billion
the odds of winning the powerball is 1/54 million
thats basically 0, okay?
maybe not the power ball is tonight if i win you wont see me next class just so you know i will be figuring out something to do but ill be figuring it out from the bahamas |
| 22:30 | and ill be thinkin gof all of you when x is very very large, this approaches 0 alrigt now what happens when x approaches from the plu side well now there putting very small numbers in the denominator so like wha is 1 over .00000001 thats 100 thousand so as the bottom gets closer and closer to 0 the whole thing gets pretty large |
| 23:01 | so as you approach 0 from this side you get positive infinity similiarly what i like to say in the math world what happens when x approaches negativ infinity, its 0 because you get a huge denominator and the whole thing gets very small and you get 0, qhat happens when you approach 0 from the minus side you get 1/ very large number but its a very large negative number so its a small number so if you had 1/negative .00001 |
| 23:31 | you get negative 100 thousand so the limit as x appproaches 0 minus side of that will be minus infinity thats why were getting this you should store this in your head somewhere cause we like to do this kind of stuff keeps you on your toes, helps bring the class average down just kidding or that kind of stuff how we doing on the whole limits to infinity |
| 24:01 | plus infinity minu infinity we sorta of understand this at some point okay time to learn something new, you got a question raise your hand because were leaving limits behind now not a lot of hand raising going on here that means you get it or your basically focusing on chemistry or something like that or candy crush |
| 24:46 | so now we gonna discuss one of those important conceptual topics of calculus that drives people crazy, there certainly will be an exam question on this okay so we gonna talk about whats call continuity |
| 25:02 | so we talk about, its very important in calculus to find functions a lot of whats going on requires a function to be continuous so what do when by continuous a continuous function is a function that doesnt have any breaks or holes in it, so if you were drawing it you never have to take the pencil off the paper how would we defind that mathematically, so a continuous function you know jut like that |
| 25:31 | where a discontinuous function stops at some point and then picks up again you dont have to copy that but continuous means theres no holes in it, in a certain region there doesnt have to be no holes everywhere but if we say in some reason and its continuous in that region there no place where you take the pencil off the paper or the pen off the paper or the chalk of the chalk board so its smooth, it doesnt have hole |
| 26:00 | it doesnt have breaks, it doesnt have asymptotes or any of those things, so how would we defined that well lets talk about the different types of breaks we can get we can get an asymptote, so we can draw something does somethin glike that its not continuous, right here whatever that number is a theres a break, so that function is discontinuous this is either called and infinite or essential discontinuity |
| 26:35 | depends on where you went to school wither word is okay theres nothing you could do to fix that it is just not continuous it continuous everywhere else |
| 27:15 | alright a second type of discontinuity looks something like that the function stops at some place and then picks up somewhere else |
| 27:32 | and it jumps, so this is called a jump disconuity and again there is nothing you can do to fix that its just a hole its a jump and you cant,what i mean by fix youll see in a minute theres nothing you can do to fix the function to get rid of that problem so your continuous over here and your continuous over here |
| 28:00 | but youre trying to get from here to here you have a discontinuity problem one thing that happens in calculus a lot of rules only apply if the function is continuous in some spots in some range, if its not continuous then we cant use the funtion, we cant use the rule third type of dicontinuity looks like that that is called a whole |
| 28:31 | just a hole, a hole in a function or a rul of discontinuity so this one, is called a rule cause you can fix the problem kind like this is a pot hole in the road, you can fill a pothole however if it looks like this or looks like that, i suggest you take a different road |
| 29:03 | this road is fixable, when it gets fixed depends on what the town is in the mood for what do these kind of things look like functionally everyone go the concept? these are the three main concepts of discontinuity either there just is not function, no value of the function theres a change in the value of the function or theres a missing value of the function |
| 29:34 | and if you think about this whenyou think about something like this, what do we know about the limit, the limit does not exist or we look here and the limit doesnt exist at a because the left and the right side disagrree but here the limit does exist at a f of does not exist but the limit exist |
| 30:06 | for example suppose this is the value like 2 this is a value like 5 then if you do the limit as x approaches a from the left side you get 2 the limit as x approaches a you get 5 since 2 does not equal 5 the limit does not exist here however the limit as x approaches a on the left side is 2 and the limit as x approaches the right side is 2 |
| 30:32 | therefore the limit exist you do not need f of x to exist f of a to exist, you only need both sides of the limit to exist so for continuity all we have to do is figure out how to get around that problem so we do a little erasing |
| 31:08 | notice what goes on here the limit as x approaches a from the minus side of f of x is l l could be anything you want the limit as x approaches a from the plus side |
| 31:36 | also equals L
so far so good, so therefore
the limit when x approaches a
of f of x
equals L
and you say yes this function is not continuous
cause theres a hole right there
you can plug the hole, so what do i need?
|
| 32:00 | i need f of a to equal L so if both of these things aree true if the limit of f of x is L as x approaches a the limit is f of L and f of a equals L then the function is continuous so we will get a little more technical in a second but this is what we are looking for |
| 32:30 | were looking for the left side limit is the same as the right side limit and that theres actually a value at a and that that value equals l becaue if the value was up here then it wouldnt be continuous so if you had a function |
| 33:06 | you have a function like that the limit as x approaches a from the plus side and the minus side is 2 so the limit as x approaches a is 2 but f of a is 5 so its not continuous so you would need both the function and the limit to have the same value so theres a slightly more technical verzison of this which |
| 33:32 | ill right down later, its in the book but this is basically what we are looking for, theres 3 things we have to figure out so lets do an example |
| 34:04 | suppose we have something like this, this is a piecewise function because its defined in pieces so the question is, is this continuous can you try and do this without graphing |
| 34:35 | what is the limit as x approaches 2 from the minus side the limit as x approaches 2 from the minus side will be values less than 2 1.99999 when i plug in a value less then 2 i get 4-1 less than 3 what happens when i plug in a value just greater than 2 |
| 35:04 | when i plug in something just greater then 2, i get omething at this part of the curve 6+5 is 11 these two do not agree that means that that function is jumping its doing something like that |
| 35:31 | so the limit from the left
doesnt equal the limit from the right, theres no limit there
so this function is not continuous at x equals 2?
just so you may know you might need to write this at some point on your exam polynomials are continuous everywhere so you can look at this and say, this is a polynomial so its continuous everywhere less than or equal to 2 this is a polynomial so its continuous everywhere greater than 2 |
| 36:02 | so your only problem is at 2 and at 2 the limit does not exist so therefore the function is not continuous alright lets take a slight variatiion on this |
| 36:47 | now we have almost the same function we have 3x+5 when x is greater then 2 and 2x+7 when x is less than 2 so is this function continuous |
| 37:02 | okay lets figure it out, well the limit when x approaches 2 from the minus side of f of x is 4 plus 7 is 11 and the limit when x approaches 2 from the minus side fo x 6 plus 5 is 11 so notice what youre really doing youre plugging 2 into both of the equations |
| 37:30 | and see if you get the same number and look
you get 11 both times so the limit
wen x approaches 2
of f of x is 11 so is it continuous at 2?
no why not? well whats f of 2 notice i wrote greater than 2 and less than 2 did not write equal to 2, i did that on purpose |
| 38:03 | remmeber your gonna take the exam at 8:45 at night so you want to make sure you catch those things so there is no f of 2 so once again that function is not continuous at f equals 2 continuous everywhere else but not at 2 okay slightly more annoying |
| 38:30 | that satisfies the first criteria when the limit existed if there wasnt a second criteria, were not quit there lets do a more annoying one |
| 39:12 | so what is the limit as x approaches 2 from the minus side when i plug in 2 i get 4 plus 7 is 11 and whats the limit ofx approaches 2 from the positive side its equal to 11 |
| 39:32 | so therefore the limit when x approaches 2 f of x is 11 but f of 2 is 10 so we satisfied almost everything we need, the limit exist f of 2 exist but they dont equal eachother this is an example of the type where you have a hole |
| 40:00 | and then a little dot in another spot it looks something like that when you get to 2 the limit is 11 but the function is only equal to 10 at exactly 2 when x is less than 2 you come from the left sde when x is |
| 40:33 | greater than 2 you come from the right side cause less than 2, greater than 2 now lets do one that works a new function so you dont get bored |
| 41:13 | suppose we put this on the test and we said is f of x continuous at x equals 2 well first thing you say is f of x is continuous everywhere other than x equals 2 because they are polynomials polynomials are continuous everywhere your only spot |
| 41:33 | cause its right where the equal shifts alright i we plug 2 in the top of the equation, well lets get the limits heres how you check, first in your head you take 2 you plug it in and you get 4 plus 3 is 7 2 and plus it in you get 12 minus 5 is 7 you get 7, 7 they agree, theres no missing equal sign so these things are going to be continuous, so how are we gonna show this |
| 42:00 | so professor and ta give you full credit you say the limit as x approaches a from the minus side of f of x thats this branch is 12 minus 5 is 7 and the limit when x approaches 2 from the plus side of this branch is also 7 therefore |
| 42:32 | the limit of f of x is 2 equals 7 and you say f of 2 equals 7 therefore its continuous at x equals 2 that would be a good enough answeer |
| 43:00 | and by the way if you want to you can abreviate continuous
cts
so you can save yourself from writing out a long word and replacing it with a short word
math is filled with that, thats why we dont write therefore we put the 3 dots
does anyone know where that came from? if i made something up youd believe it
okay so far so good?
because this is where is |
| 43:31 | where x equals 2 when x equals 2 i just plug in 2 and i get 4 plus 3 is 7 rememebr on this one, i dont have a problem where x equals 2 but here i do and there i do so watch for that, less than greater than or equal to sign thats one of the ways we get you guys |
| 44:00 | so lets see, whats a typical type of question alright well give you guys one to work oj |
| 44:54 | find the value c so f of x is continuos on the interval minus infinity |
| 45:00 | plus infinty, you want to know why you write those with perenthasis and not square brackets because you cant actually have infinity, you cant actually have a square braket dont use a square bracket, use a parenthesis find the value c where f of x is continuos everywhere c is cx squard plus 4x when x is greater than or equal to 3 and 5x plus c when x is less than 3 |
| 48:20 | alright thats long enough so in order for this to be continuous the limit from the left side has to equal the limit from the right side |
| 48:30 | and they both have to equal f of 3 so the limit from the left side well we say when x is less than 3 5 times 3 is 15 plus c and the limit from the right side is you plug in 3 |
| 49:00 | and you get 9 c plus 12 thats also what f of 3 equals by the way so in order for these to be continuous this has to equal that so 15 plus c has to equal 9c plus 12 so you do a little algebra and you get is 3/8 |
| 49:34 | well i may made a mistake you never know you take 3 and you plug it in here c times 3 square is 9, so 9c 4 times 3 is 12 alright lets give you one thats more annoying |
| 50:56 | alright we have 3 parts of this function |
| 51:00 | we have x squared minus 4 over x minus 2 is less than 2 ax squared minus bx plus 3 when x is between 2 and 3 and 2x-a+b when x is greater than or equal to 3 find a and b to make this continuous everywherre good exam question in order for this to be continuous the limit as x approaches 2 from the minus side has to equal x minus 2 from the plus side |
| 51:33 | so the limit x approaches 2 from the minus side of f squared minus 4 over x-2 problem is you plug in 2 you get 0/0 to think about this and say wait i can factor the top |
| 52:00 | since its not actually 2 we can cancel this and get 4 so the limit as x approaches 2 from the minus side of this equation is 4 and what about the plus side that equals a times 2 squared so 4a minus b times 2 |
| 52:31 | so this limit has to equal this limit, has to equal f so 4a minus 2b plus 3 has to equal 4 4a minus 2b has to equal 1 so far so good now lets do the same as x approahes 3 limit of x approaches 3 from the minus side |
| 53:02 | is the middle function now were plugging in 3 not 2 and the limit as x approaches 3 from the plus side of f of x equals 6 minus a plus b so when we plug those together we get 9a |
| 53:37 | 3b plus b has to equal 6 minus a plus b so 10a minus 4b has to equal 3 so there you go we got 2 equations and 2 variables so we got 2a minus 2b equals 1 |
| 54:00 | and 10a minus 4b equals 3 a is -5/2 lets see what id do, id multiply this equation by negative 2 and i would get negative 8a plus 4b is negative 2 made a mistake there and 10a minus 4b equals 3 |
| 54:31 | and you get 2a equals 1, a equals 1 a excuse me now that i know what a is i can go bac and find what b is see what i did, should be able to solve 2 eqautions and 2 variables now that i know its a half i know 10 times 1/ minus 4b equals 3 so minus 4b equals minus 2 |
| 55:01 | b also equals a half so i add these 2 equations together i get minus 8a and plus 4b plus 4 b and minus 4b cancels well you can do substitution but substitution is messy you can solve this however you want what did i do here, i took this equation |
| 55:31 | i multiply it by negative 2 so that becomes 8a minus 2 times minus 2 is 4b minus 2 times 1 is -1 and i took the other equation and put it underneath now i add them together 10 minus 8 is positive 2 the b's cancel and then you get 1 2a is 1 thats the methond of symetanious equations like i said did memories from 9th grade that was a long time ago for some of you guys |
| 56:00 | longer though for me 9th grade was a while ago i can only count that high its an limit approaching infinity but thats some math word do you want to know why i used negative 2 theres a bunch you couldve done but the easiest you could do,a another you could do you could divide 2 by something and you get b |
| 56:31 | and you get b alone, you have a lot f techniques
to decide at that point
everyone understand what i did
sould i do one more like this?
who votes for 1 more who does not vote, whos had enough thats a tough call, youve had enough lets do one more |
| 57:04 | ill make it slightly easy |
| 57:50 | okay why dont we do this as a team so we have this really great function |
| 58:02 | ax+b-1 when x is less than 1 x squared minus 3ax-2b when x is greater than 1 but less than 2 and 4ax plus b plus 3 when x is greater than or equal to 2 do you know where these equations show up? they show up in like theyll show up when you have a change going on for example a business function youll have cost or expenses and |
| 58:30 | this is if youre running 1 machine this is if youre running 2 machines this is if your running more than 2 machines things like that so you see this a lot in operations i cant speak for biology cause i took biology in 9th grade and 9th grade was not recently i took biology from about a mile from here i took calculus here |
| 59:03 | alright so whats the limit when x approaches 1 from the minus side well plug n 1 in the top branch and you get a plus b minus 1 the limit when x approaches 1 from the plus side go to the middle branch and you get 1+3a-2b |
| 59:31 | these have to equal eachother because as you approach 1 from both sides you have to get to the same spot a+b-1 has to equal 1+3a-2b do a little magic algebra and you get 2a-3b equals -2 alright now lets repeat for the bottom 2 things |
| 60:03 | the limit as x approaches 2 from the minus side you plug in 2 you get 4 plus 6a-2b and when you appraoch 2 from the plus side |
| 60:32 | you get a a plus 2 plus 3 and these have to equal eachother so 4 plus 6a minus 2b has to equal 8a plus b plus 3 so 2a plus 3b has to equal 1, wow that came out well didnt it |
| 61:05 | really proud on myself, so 2a minus 3b minus 2\ and 2a plus 3b equals 1 i hope you can solve this you simply add these two equations together you get 4a equals negative 1 a is negative a qauter |
| 61:42 | howd we do?
so now you got a as a qaurter so now you go back and find b so 2 times a quarter plus 3b equals 1, 3b is a haldf |
| 62:02 | b is 1/6 |