The value of $\lim _\limits{\substack{x \rightarrow 1 \\ y \rightarrow 0}} \frac{y^3}{x^3-y^2-1}$ as $(x, y) \rightarrow(1,0)$ along the line $y=x-1$ is-

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

The value of

\lim \limits_{\substack{x \rightarrow 1 \\ y \rightarrow 0}} \frac{y^3}{x^3-y^2-1}

(x, y) \rightarrow(1,0)

along the line

y=x-1

is-

View answer and explanation

Correct option: A

Option A (0) is correct because substituting y = x - 1 and taking the limit as y approaches 0 results in a value of 0.

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