Limit of Multivariable Functions | Calculus (In English)

Hello friends, in this video we are going to discuss limit of multivariable functions. First, we begin with the definition let f of x y be a function of two variables x and y. The limit of f of x y as x y approaches to a b is l if for given epsilon greater than 0 there exists delta greater than 0 such that square root x minus a whole square plus y minus b whole square less than delta implies f of xy minus l modulus less than epsilon. The notation for the limit is limit xy approach to a b f of xy is equal to l. Now there is a two types of problem in limit of multivariable function. First is a problem of existence of limit and second is a problem of non-existence of limit. So one by one we discuss both type of problems first we are going to discuss the problem of existence of limit example one if f of x y equal to x square y square divided by x square plus y square show that limit x y approach to 0 0 f of x y equal to 0 to prove the limit for given epsilon greater than 0 we want to find delta greater than 0 such that these two inequality will satisfy by epsilon and delta so here we want to find delta for the given epsilon square root x minus a whole square plus y minus b whole square less than delta where a is equal to b is equal to 0 0 which is a point at which we are going to find the limit and l is the limit value which is 0 so we have square root x minus 0 square plus y minus 0 square less than delta implies f of x y minus 0 less than epsilon where l is a value of limit Now, in the simplified form square root x square plus y square less than delta implies modulus f of xy minus 0 less than epsilon. So, we begin with the second part f of xy minus 0 is equal to f of xy is x square y square divided by x square plus y square minus 0. Then, we have modulus x square y square divided by x square plus y square. But we know x square less than equal to x square plus y square and y square less than equal to x square plus y square we use this relation because we try to convert this function in the form of x square plus y square therefore we have less than equal to x square plus y square x square plus y square divided by x square plus y square 1 x square plus y square cancel we have x square plus y square now we know x square root x square plus y square less than delta therefore x square plus y square less than delta square so if we take delta square is equal to epsilon or delta equal to square root epsilon then we have f of x y minus 0 less than epsilon which we required so if square root x square plus y square less than delta then we are able to find f of x y minus 0 less than epsilon and hence limit x y approach to 0 so f of x y equal to 0 so if limit exists then we'll use that epsilon delta definition now we're going to discuss the problem of non-existence of limit if f of x y is equal to x y square divided by x square plus y power 4 show that limit x y approach to 0 0 f of x y does not exist we know that limit has always unique value therefore for every path from x y approach to 0 0 limit should have same value so if there exists at least two different path from x y to 0 0 for which limit will have two values then limit x y approach to 0 0 f of x y does not exist so it means if we are able to find two path from x y to 0 0 for which limit has different values then limit cannot exist because limit is always unique as we know we have infinitely many path from x y up to 0 0 so first question how can we select the correct path we will use two types of path x is equal to m y power k and y equal to m x power k so here first question which variable will be the subject of the equation second question what is the value of power k so for the first question first we identify the variable which has smaller power in the function just for example x y square in the numerator and the denominator x square plus y power 4 so we can see x is a lower power so the first path where x is subject will be the answer so now we know the x is equal to m y k should be the path now what is the value of k for that we take the ratio of power of x and y so in the numerator ratio power of x and y so it is 2 divided by 1 so which has value 2 and 4 divided by 2 which has also value 2 so numerator and denominator will give the same ratio value 2 2 so k equal to so your path is x is equal to my power now limit x y approach to 0 0 f of x y equal to limit x y x y square divided by x square plus y square now substitute x is equal to my square so my square y square my square square plus y power 4 now this whole function convert into y so we have limit of y y approach to 0 now if we simplify then we have m y power 4 and the denominator m square y power 4 plus y power 4 so from the numerator and denominator y power 4 cancel then we have final form of the limit m divided by m square plus 1 now we know m is a parameter so we can take the different values of m so we'll get the different path so if we take m equal to 1 we have path 1 x is equal to y square m equal to 1 so 1 upon 1 square plus 1 so 1 divided by 1 plus 1 equal to 1 by 2 so for path 1 limit as value 1 by 2 m equal to 2 path 2 x is equal to 2 y square then limit as value 2 divided by 2 square plus 1 so which is 2 by 5 so limit has two different values for two different path therefore limit does not exist now these are the examples and this example we will see which path is proper for the non-existence problem so limit x y approach to 0 0 x y divided by x square plus y square so here both the variable have same power so we can take any path x is equal to or y equal to so x equal to y power because ratio 1 upon 1 is 1 2 by 2 equal to 1 so ratio will 1 so path is x is equal to my or y equal to mx now this example we already discussed which has path my square now next x square y divided by x power 4 plus y power 2 so here y is a lower power therefore y is a subject of the equation m x power 2 divided by 1 4 divided by 2 ratio is 2 so m x power now next example x y cube divided by x square plus y power 6 so naturally x should be the subject of the equation because it has a lower power ratio is 3 divided by 1 which is 3 and 6 divided by 2 which has value 3 so power of y is 3 so x m y cube now next example x square minus y square divided by x square plus y square so same way both have the same power so x is equal to m y and y equal to m x because 2 divided by 2 is 1 and 2 divided by 2 is 1 so this way we can select path for the non-existence problem thanks for watching for more videos like share subscribe press the bell icon of next gen maths