2 votes 2 votes What is value of $\lim_{x\rightarrow 0, Y\rightarrow 0}\frac{xY}{x^2+Y^2}$ 1 -1 0 Does not exist Calculus limits + – minal asked Nov 27, 2016 • edited Nov 28, 2016 by minal minal 2.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $\lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^2+y^2}$ = $\lim_{x\rightarrow 0 y\rightarrow 0}\frac{1}{2x+2y}$ =$\frac{1}{2.0+2.0}$ =$\alpha$ Ans D) srestha answered Nov 27, 2016 srestha comment Share Follow See all 9 Comments See all 9 9 Comments reply minal commented Nov 27, 2016 reply Follow Share you just changed the question , at numerator there is xY not xy ... 0 votes 0 votes Arjun commented Nov 28, 2016 reply Follow Share what is Y? @srestha if the limit is $\infty$ do we say limit does not exist? For this LHL != RHL must be there rt? 0 votes 0 votes minal commented Nov 28, 2016 reply Follow Share sorry sir its Y at all y places ... and limit also atY.. and x and Y both are diffferent variables .. 1 votes 1 votes srestha commented Nov 28, 2016 reply Follow Share @Sonam I actually edited because some latex shown in ur question, so the question was hard to read Actually Y or y does it matter? It is a variable only 0 votes 0 votes srestha commented Nov 28, 2016 reply Follow Share @Arjun Sir u mean to say $\lim_{(y/x)\rightarrow \alpha}$ But, in some cases $\lim_{z\rightarrow \alpha }$ also possible So, I think we cannot directly say like this. Though if u can prove it, u can take a try :) 0 votes 0 votes Anusha Motamarri commented Nov 28, 2016 reply Follow Share @srestha.. u got infinity... how can we say that limit doesnt exist? @arjun sir, x-->0 and y-->0 dont we need to check 4 possibilities? x and y approaching zero from right side x and y approaching zero form left side x approaching zero from left side and y approachng zero from right side x approaching zero from right and y from left i think..if any two of these 4 arent equal we shud say thay limit doesnt exist. bcoz from any side we approach to zero we shud get the same value. then only we can say limit exists. this will be a graph formed with x axis and y axis.. we need to check whats happening around the origin. plz verify sir. 3 votes 3 votes minal commented Nov 28, 2016 reply Follow Share @srestha yes ,no difference , earlier i wrote wrong question ..now correct but m not getting ur method 0 votes 0 votes srestha commented Nov 28, 2016 reply Follow Share @Anusha Actually it is asking about value of limit infinity means no real value could exists rt? In ur approch how do u prove LHL!= RHL ? 0 votes 0 votes Pavan Kumar Munnam commented Nov 28, 2016 reply Follow Share @srestha if u get infinity it doesnot mean that limit doesnot exit...but it exits with a value of infinity(undefined) where as ur right limit give you +infinity and left limit gives u -infinity then it is doesnot exist 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes Answer is C ? Pavan Kumar Munnam answered Nov 28, 2016 Pavan Kumar Munnam comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Sushant Gokhale commented Dec 3, 2016 reply Follow Share @Sreshtha. Even I think (C) should be the answer. Lets see why. $\lim x\rightarrow0 y\rightarrow0$ $\frac{xy}{x^{2}+y^{2}}$ = $\lim x\rightarrow0 y\rightarrow0$ $\frac{xy}{x^{2}+y^{2}}$ = $\lim x\rightarrow0 y\rightarrow0$ $\frac{x}{x^{2}+y^{2}}$ + $\lim x\rightarrow0 y\rightarrow0$ $\frac{x}{x^{2}+y^{2}}$ .........Y times Now, $\frac{x}{x^{2}+y^{2}}$ tends to 1. ........................(1) Also, we are taking this sum Y times. Let , Y = 1/Z and let, Z -> infinity Thus, we are dividing (1) by infinity. Thus, this tends to 0. Dont know if right approach or wrong. 0 votes 0 votes srestha commented Dec 3, 2016 reply Follow Share how, y=1/z if u do it, change whole equation in z 0 votes 0 votes Sushant Gokhale commented Dec 3, 2016 reply Follow Share yes. we can. If y->0, then replace y by 1/z such that z->infinity 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes The question can be changed to (1)/(x/y + y/x) , if y tends to zero on a line y = mx then the limit = (1)/(m+1/m) which is not a fixed value.So a particular limit does not exist. Sai Shravan answered Jan 28, 2019 Sai Shravan comment Share Follow See all 0 reply Please log in or register to add a comment.