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 2.2k 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 Show 6 previous comments 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.