hello everyone welcome to the next lecture on the calculus today we will discuss about how you can calculate the limits of the several variables myself dr girl working in the school of mathematics harper institute so as we all know that how you can compute the limits of a one-dimensional function so we all know that how you can compute this we can take the limits from the zero minus and then we will compute the limit as a zero plus and check whether they are equal or not if they are equal then we can say the limit exists otherwise not but the question arises is how you can find the limits of more than one variables like of here there are the two variables x and y and so on so that means the objective of this lecture is to explain how you can find the limits of the several variables so i will discuss about the two-step approach what are the two steps are there we will firstly choose the path and the direction and then we will find the limits are there if this limit is exist but not unique then the limit does not exist so remember if it is not unique then limit does not exist but if the limit is unique then we can't say that whether the limit exists or not look at the couple of the examples in in this presentation so what is the concept behind is that we all know that how you can compute the limit as x approaches a of the f of x how you can compute them we will take the limit as a zero minus a minus that is we have to taken all those points which are taken from this direction and then we will take the limits as a a plus that is we are direction from right hand side that is the right hand limit and this is the left hand limits are there but and we will check whether this both our limit exists and unique then we can say the limit existence is there and moreover it should be the finite in nature but that is we will take the limits either from the left hand side or from the right hand side however if you are taking this two dimension picture like of here how how you can reach at this point either you can start from you can reach through the straight line either you can go from this side or either you can start from here or you can go from like this way so there are so many ways are there which you can reach at this point zero comma zero however in the one dimension you all know that this is only the real line so you can reach at the point a either through the left side or to the right side that's why there are only the two ways are there but in in these 2d dimensions so you your targets you reach at the point 0 0 how do you reach that you can go either on this left side or right side or you can go from here or even you can start from here you can go this side and then go this side so there are so many ways out there so the question arises is how you can approach towards this point why this is 0 0 because i have taken the limit as 0 0 or in journal there are how you can approach at this point x 0 y 0 either you can approach this through this straight line or you can approach through this curve or you can go like this way and here and many more are there that is either you can use as a straight line either you can use as any of the path or either any of the curve is there so our target is to check the limit on each of the way and if you find that the limit does not exist to any of the direction like say if the limit does not exist any of this any of this path then we can say the limit of the two variables does not exist and this test is called as the two path test for the non-existence remember does not exist means the limit is not existence however if this limit exists and the unique then you can't say about the existence of the limit because this test is applicable only for the known existence we will see in the couple of examples so we uh when you can say this limit exists when you will prove that the limit exists for this path limit exists for this part limit exists for this curve and all are there so it will be exist until for all these possible path or the direction so as i discuss we will discuss the two steps rule which we have explained with the help of the 12 numerical examples so there are the two categories which i divided into them the first way is where you have to check whether the limit exists or the something question around whether evaluates the limit second part is whether you have to show proof that the limit of this is here like of this you have to prove that limit is my zero you have to prove that here on the other hand evaluate the limit check whether the existence of the limits or not so there are the two categories are there so let's say we will discuss the 12 examples are here so check the existence of here so it means this is my type one so whenever there is a type one in this category we will choose the path how you can choose the path that's a very simple i will tell you the trick are there how you can choose the path are there you can take this denominator part as a zero and then find the value of the y from here so this is my here now forget about this plus minus sign we will take only the magnitude value so this is my here so this is the curve are there so i can take as a m where m is any scalar value our target is to check this value so i can take this path here now i can substitute this value of the m in everywhere here and take the limits as 0 comma 0 so i can substitute the value of d y here so it will be this one so you can see this is the x raised power 12 will be common it will be cancel out here now it is independent of the x so the answer of this will be my here but since m is my real number so it can take any of the values so what is the meaning of that if you take m is 1 the answer will be my 1 by 8 if you take m is 2 the answer will be my 2 by 27 and so on so it means limit is not unique so since it depended upon the amp so limit is not unique hence the limit of this function does not exist so this is my type one look at one one more example again you have to check whether the limits are there or not so again this is a type one so what is the path is you can take the denominator part as a zero so what's the value of d y is x that is m of x so you can take m is any of the real numbers so i can take the path that is a curve so we can take the line as y is equal to mx substitute y is equal to mx in here you can take x will be common it will be cancel out you will get limit as a zero so since this limit exists and it's a unique but can you say the limit existence no because if it is exist this is we check only for y is equal to mx but remember that the limit should be existing for all the path it means we have to check whether it is existence for here whether it is existence for this also and so on so it is not an easy task for that so we can't say that the limit of this function exists right now because we we have we don't have enough time to check all these possible paths so we can so this test does not conclude whether the limit exists or not so that is the remark for you that is the two path test is used only for the known existence of the limits and whenever you want to calculate the limits so we will use this type 2 how you can compute that whatever the limit you want to calculate what we can do is we can start with this definition we can take in an epsilon that's a very small number such that x minus x 0 x minus x 0 is is a very small number y minus y 0 is a very small number we can start with this lhs side and our targets to prove whether it is less than of the small number or not if it is so then if it is exist then the limit is always existence it remember the always otherwise limit does not exist so look at this example since this test does not give us any of the conclusion so we can start with this one this is my l because the limit is my l we can start from here so let epsilon be greater than 0 we are given such that x minus x 0 here is x 0 is my 0 y 0 is my 0 i take less than so i can substitute the value of the fx is given to be here i can substitute here is this one now we all know that what is the value of the mode of x y it is nothing but my what is the mode of x y it is nothing but my here now if you look about that this number is nothing but my here what is the mode of the x minus y mod of x minus y is nothing but less than equal to mode x plus mode one now i can substitute this value it is less than of the delta this is less than delta this is so it is it is my 2 of delta square so i can consider this as upside so i can start from here it is less than epsilon so it means this limit exists and the limit of the f of x is my zero and that's the required answer of this we will see this definition in the couple of examples look at that you have to prove that it means it is a type two you have to start with the definition of epsilon let epsilon be greater than 0 such that x minus x 0 is less than delta y minus y 0 is less than 0 this is my f of x comma y this is my l so i can start with here our target is to prove is less than of the delta so i can substitute the value here we all know that what is the value of the mode x plus y it is less than of the mod x plus mode y i can write like here again this is the mode of the x y which can be written in this one what is the value what is the sign what is the mode of this sign one by it is always be less than equal to one so i can write this value as mode y plus mod x what is the mode of y you can see it is less than of the delta plus delta that is here i can consider this as upside that is here is my of this i can consider this as my epsilon so this means f of x minus f of 0 is less than epsilon whenever this two what is the meaning of that limit exist so once the limit exists so what is the value of the limit is my here this is my zero so you can say the limit is my zero is the required us now you can see this is the evaluate so it is my type 1 you have to choose the path order again you have to check here it is again this so if you consider this one firstly what is the path of this if you substitute the denominator part as a zero what is the value of this this is my m this is my minus so you can ignore the negative sign so y is nothing but my x square so i can take as m of x square so you can choose the curve as y is equal to mx square substitute the value here substitute the value as y here what is that you can see m square and x raise power 4 will be cancel it will be here so you can see for the different value of the m it has different answers so it means the limit does not exist because it is not a unique look at this one so what is the path is you can take this part as a zero so y is nothing but my x square so i can take it as m x square so i can choose the curve as y is equal to mx square substitute this value here you will get as of this clearly says that this value is my infinity if m is 1 and it will be 1 minus m if m is not equal to 1 and you can see there are not a unique answer are there moreover for the different value of the m it has the different answers so it means again the limit is not unique so hence the limit does not exist check whether limits are there so again it's my type one you have to choose on the path like say for example i can consider this example firstly you can choose denominator as a zero you can consider this path substitute this value here you can see x will be cancel out you will get this expression and again it is not unique because it depends on the value of the amp for the different value of m they have the different answer limit does energies for here you can take the curve this is 0 y 6 is nothing but x square y is my 1 by x x raise to power 1 by 3 so i can take as m x raised to power here or y cube is my so you can substitute this y cube as here and after the calculation you will see it is again dependent on the m so the limit is not unique hence limit does not exist this is again the evaluates are there so you have to consider this is my type one again this is evaluate so it is my type is still so how you can do that if you start from here you can take that denominator part as a 0 so you can take this value as of root y so i can taken is as m of root y or you can find the value of the y from here it is my uh it is not possible to find the value of i so i can take this y from here is m root y minus of x or you can take like here so i can substitute the value of the x plus y wherever x plus y is written here here and here you can see this one now while taking y is equal to 0 it what's the answer of this it will be 0 but you can see there are the two answers if 0 only if m is not equal to plus minus 1 because f whenever m is plus minus 1 it becomes infinity so if m is plus minus 1 it means there are the two different answers are there again limit is not unique hence the function the limit of this function does not exist how you can take this part you can take this value as a zero now you can take this value as of say m of x cube you can find the value of the y mx cube plus of 3x you can take this curve as of here that is 2x minus 3 is nothing but mx cube so i can substitute this value here and i can find the value of the y from this this is mx cube plus 3x upon 2 i can substitute the value of y here by here now after this you can see x cube will be cancel out whenever x will be 0 this part will be 0 so the answer is 9 by 4 of 1 plus m again it depended on the value of the m so therefore limit does not exist again both the questions are the type one because they are one to evaluate them how you can solve this you can take this part as a zero so what you can do that y cube minus x is nothing but my x cube i can taken it as a m or y cube is nothing but my x plus m cube or you can also take this as x minus x cube and then y is nothing but my m x minus x cube of this so that's is your on purpose r so you can take this curve as y minus x cube is mx cube on this side you can take y 4 minus x is nothing but mx4 so if you substitute this value as a here this value as a here and after the calculation you can see this value is dependent upon the m this value is also dependent on the m hence a limit does not exist now evaluate the limit again so here you want to evaluate the limit so again it is my type 1 what you can do that you can simply take x minus y of 0 is this one x is nothing but my y so that is y is equal to mx or you can consider this value as x square y square is x minus y whole square so what is the meaning of that xy is nothing but my x minus y so you can consider this as m so you can find the value of the y from here it is nothing but my mx plus 1 so that's is depending upon u so if you substitute this value value as here you can see when you take x is 0 so it's a 0.0 is a 0 but if m is not equal to 1 what will happen if m is 1 if m is 1 it will be zero it will be my one so you can clearly say there are the two different answers corresponding to the different value of the m it means this uh sorry this is my one so once it will be there so you can see the limit does not exist hence the function limit of this function is not with it now this is my type 2 because you want to prove that again this is my type 2 so it means you start with the value of the epsilon greater than you so let epsilon be greater than 0 such that x minus x 0 is less than delta y minus y 0 is less than delta and this is your target we can start from here we can substitute the value of the f x and here now what you can do that this value mode of x is nothing but my delta this value is nothing but delta and you all knows that for all the values of the x and y this number is always be less than 1 why because x square minus y square is always be less than of x square plus y square so this number is always be less than 1 so you can write here so this value is nothing but my delta square i can consider this as upside so this value is less than of the epsilon hence limit exists and the answer is my 0 similarly for here we can take epsilon as this we can substitute this value but is a mode of x plus y we can write this as software sine value is always be less than of the one you can write like this way what is that i can consider this as mod x and i can write like of this way x square plus y square we can see this is mod x this is mod x i can write this value as of can i write this value like of this and we all know that this value is always be less than one so next line i can write this value as of this this value is delta this value is delta i can consider this as upside so that's a required proof of this i can consider this as a one so this is my two delta it is epsilon hence this value exists so therefore limit is here so that is all about this all 12 plus questions are there i hope you can simply enjoy this type one and type to both based on this limit of the function we can define the continuity of the two variable in our next class till then you can simply like share and comments on this video best of luck students happy learning