Question

$\underset{\genfrac{}{}{0ex}{}{x\to 1}{y\to 0}}{\lim }\frac{y^3}{x^3-y^2-1}$ as $(x,y)\to (1,0)$ along the line $y=$ $x-1$ is given by

• A. 1
• B. $\infty$
• C. 0
• D. None of these

Answer 1

Answer: (C)

Since $y=x-1$,
$\therefore x=y+1\text{. }$
As $(x,y)\to (1,0)$ along the line $y=x-1,x=y+1$ holds throughout.
$\therefore \underset{\genfrac{}{}{0ex}{}{x\to 1}{y\to 0}}{\lim }\frac{y^3}{x^3-y^2-1}$
$=\underset{y\to 0}{\lim }\frac{y^3}{(y+1{)}^3-y^2-1}$
$\underset{y\to 0}{\lim }\frac{y^3}{y^3+2y^2+3y}$
$=\underset{y\to 0}{\lim }\frac{y^2}{y^2+2y+3}=\frac{0}{3}=0$