Question

$\displaystyle\lim _{x \rightarrow 1 \atop y \rightarrow 0} \frac{y^{3}}{x^{3}-y^{2}-1}$ as $(x, y) \rightarrow(1,0)$ 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 $(x, y) \rightarrow(1,0)$ along the line $y=x-1, x=y+1$ holds throughout.
$\therefore \displaystyle\lim _{x \rightarrow 1 \atop y \rightarrow 0} \frac{y^{3}}{x^{3}-y^{2}-1}$
$= \displaystyle\lim _{y \rightarrow 0} \frac{y^{3}}{(y+1)^{3}-y^{2}-1}$
$ \displaystyle\lim _{y \rightarrow 0} \frac{y^{3}}{y^{3}+2 y^{2}+3 y}$
$= \displaystyle\lim _{y \rightarrow 0} \frac{y^{2}}{y^{2}+2 y+3}=\frac{0}{3}=0$