Q 13. - Calculus

Question

Show that when C is either the $x- or y-axis$, we have $\underset{\underset{C_{}}{(x,y)\to (0,0)}}{\lim }\frac{x^{2}y}{x^4+y^2}=0$.

Step-by-Step Solution

Answer

It can be shown that limit is zero at both $x=0 and y=0$.

Step 1: Given Information

Consider that the limit is $\underset{\underset{C_{}}{(x,y)\to (0,0)}}{\lim }\frac{x^{2}y}{x^4+y^2}=0$

 The objective is to prove that the limiting value is $0$, when C is either the $x- or y-axis$.

Step 2: Evaluating the limit 1

Assume that curve C is on the $x-axis$, with $y=0$. Calculate the function limit as you approach point $(0,0)$ on the curve C as follows: 

$\underset{\underset{C_{}}{(x,y)\to (0,0)}}{\lim }\frac{x^{2}y}{x^4+y^2}=\underset{\underset{}{\left(x\right)\to \left(0\right)}}{\lim }\frac{x^2\left(0\right)}{x^4+{\left(0\right)}^2}=\underset{\underset{}{\left(x\right)\to \left(0\right)}}{\lim }\frac{0}{x^4}=0$

Step 3: Evaluating the limit 2

Assume that curve C is on the $y-axis$, with $x=0$. Calculate the function limit as you approach point $(0,0)$ on the curve C as follows: 

$\underset{\underset{C_{}}{(x,y)\to (0,0)}}{\lim }\frac{x^{2}y}{x^4+y^2}=\underset{\underset{}{\left(y\right)\to \left(0\right)}}{\lim }\frac{{\left(0\right)}^{2}y}{{\left(0\right)}^4+y^2}=\underset{y=0}{lim}\frac{0}{y^2}=0$

Hence, the limiting value of the function $\frac{x^{2}y}{x^4+y^2}$ when $(x,y)\to (0,0)$ along the curve C, here, C is either the $x- or y-axis$ is $0$ .