VALUE OF LIMIT

Question

What is value of  $\lim_{x\rightarrow 0, Y\rightarrow 0}\frac{xY}{x^2+Y^2}$

• A. 1  
• B. -1
• C. 0
• D. Does not exist

Answer 1

$\lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^2+y^2}$

= $\lim_{x\rightarrow 0 y\rightarrow 0}\frac{1}{2x+2y}$

=$\frac{1}{2.0+2.0}$

=$\alpha$

Ans D)

Comments

@srestha yes ,no difference , earlier i wrote wrong question ..now correct but m not getting ur method

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@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 ?

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@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

Answer 2

Answer is C ? 

Comments

@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.

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how, y=1/z

if u do it, change whole equation in z

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yes. we can. If y->0,  then replace y by 1/z such that z->infinity

Answer 3

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.