ISI-2016-05

Question

Let 

$$f(x,y) =  \begin{cases}  \dfrac{x^2y}{x^4+y^2}, &  \text{      if } (x,y) \neq (0,0) \\ 0 & \text{    if } (x,y) = (0,0)\end{cases}$$

Then $\displaystyle{\lim_{(x,y)\rightarrow(0,0)}}f (x,y)$

• A. equals $0$
• B. equals $1$
• C. equals $2$
• D. does not exist

Answer 1

Approaching (0,0) along the line $y = mx+c$

$\Rightarrow$$\lim_{(x,y)\to(0,0)} \frac{x^2y}{x^4y + y^2}$

$\Rightarrow$$\lim_{(x,y)\to(0,0)} \frac{x^2.mx}{x^4.mx + m^2x^2}$

$\Rightarrow$$\lim_{(x,y)\to(0,0)} \frac{mx^3}{mx^5 + m^2x^2}$

$\Rightarrow$ $\lim_{(x,y)\to(0,0)} \frac{x}{x^2+ 1.m}$ $ = 0$

Approaching (0,0) along the parabola

$y = x^2$

$\Rightarrow$$\lim_{(x,y)\to(0,0)} \frac{x^2y}{x^4 + y^2}$ $ = $ $\lim_{x\to0} \frac{x^4}{x^4 + x^4}$ 

 $= $$\lim_{x\to0} \frac{x^4}{2x^4 }$

$\Rightarrow \frac{1}{2}$

THe Limit does not exist.

answer is (D)

Comments

why did you equate with line and parabola?