Evaluate the limit: lim_{(x, y) to (1, 0)} {x^3 - 1} / {x^3 + 2 x^2 y + x y^2 - x^2 - 2 x y - y^2}.

Question:

Evaluate the limit:

{eq}\displaystyle \lim_{(x,\ y) \to (1,\ 0)} \dfrac {x^3 - 1} {x^3 + 2 x^2 y + x y^2 - x^2 - 2 x y - y^2} {/eq}.

Limit Of A Multivariable Function :


The limit of a multivariable function gives us an idea about how the function behaves as it approaches the given limit.

In our problem, we have to evaluate the limit of the multivariable function. For this, first, we use the factorization method and then substitute the variables with the values of the limit.

Answer and Explanation:

Given

{eq}\lim_{(x,\ y) \to (1,\ 0)} \dfrac{x^3 - 1}{x^3 + 2 x^2 y + x y^2 - x^2 - 2 x y - y^2} {/eq}

We have to evaluate the limit.


{eq}\begin{align} \lim_{(x,\ y) \to (1,\ 0)} \dfrac{x^3 - 1}{x^3 + 2 x^2 y + x y^2 - x^2 - 2 x y - y^2} &=\lim_{(x,\ y) \to (1,\ 0)} \dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x^3+2x^2y+xy^2\right)+\left(-x^2-2xy-y^2\right)}\ & \left [ \because x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right) \right ]\\ \\ &=\lim_{(x,\ y) \to (1,\ 0)} \dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+2xy+y^2\right)-\left(x^2+2xy+y^2\right)}\\ \\ &=\lim_{(x,\ y) \to (1,\ 0)} \dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+2xy+y^2\right)(x-1)}\\ \\ &=\lim_{(x,\ y) \to (1,\ 0)}\dfrac{x^2+x+1}{\left(x+y\right)^2}\\ \\ &=\dfrac{1^2+1+1}{\left(1+0\right)^2}\ & \left [ \text{Substitute the value of the limit} \right ]\\ \\ &=\dfrac{3}{1}\\ \\ &=3 \end{align} {/eq}


{eq}\color{blue}{\lim_{(x,\ y) \to (1,\ 0)} \dfrac{x^3 - 1}{x^3 + 2 x^2 y + x y^2 - x^2 - 2 x y - y^2}=3} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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