Question Video: Evaluating the Limits of a Multivariable Function Using Factorization of Difference of Cubes Mathematics

Evaluate lim_((π₯, π¦) β (0, 0)) (π₯Β³ β π¦Β³)/(π₯Β² + π₯π¦ + π¦Β²), if it exists.

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Video Transcript

Evaluate the limit as π₯, π¦ approaches zero, zero of π₯ cubed minus π¦ cubed over π₯ squared plus π₯π¦ plus π¦ squared, if it exists.

Here, weβre dealing with a multivariable function. Itβs a function that involves more than one variable; this is π₯ and π¦. And so, we recall that if a function in π₯ and π¦ is continuous at some point π, π, then the limit as π₯, π¦ approaches that point of π of π₯, π¦ is equal to π of π, π. Essentially, if we know that a function is continuous at a point, then to find the limit of the function at that point, we just plug that point into the function. So, our job is to establish whether π₯ cubed minus π¦ cubed over π₯ squared plus π₯π¦ plus π¦ squared is continuous at zero, zero.

Now, at first glance, it might seem that it is not. If we were to substitute π₯ equals zero and π¦ equals zero into this function, we get zero over zero, which is of indeterminate form. But letβs see if we can manipulate our function somewhat. Well, the numerator, π₯ cubed π¦ cubed, is the difference of two cubes. And we know we can write this as π₯ minus π¦ times π₯ squared plus π₯π¦ plus π¦ squared. And then, our fraction becomes π₯ minus π¦ times π₯ squared plus π₯π¦ plus π¦ squared all over π₯ squared plus π₯π¦ plus π¦ squared. And we see we have a common factor of π₯ squared plus π₯π¦ plus π¦ squared. So, we divide through. And our entire fraction simplifies really nicely. And weβre simply left with π₯ minus π¦.

And so, the question weβre now going to ask ourselves is, is the function π₯ minus π¦ continuous at zero, zero? π₯ minus π¦ is a multivariable polynomial. And just like single-variable polynomials, these are continuous over their entire domain. And this means the function π₯ minus π¦ is continuous over its entire domain, and therefore it must be continuous at zero, zero. And so, to find the limit as π₯, π¦ approaches zero, zero of π₯ cubed minus π¦ cubed over π₯ squared plus π₯π¦ plus π¦ squared, weβre going to find the limit as π₯, π¦ approaches zero, zero of π₯ minus π¦. And weβre going to achieve this by letting π₯ be equal to zero and π¦ be equal to zero. This gives us zero minus zero, which is of course equal to zero. And so, we see that the limit as π₯, π¦ approaches zero, zero of π₯ cubed minus π¦ cubed over π₯ squared plus π₯π¦ plus π¦ squared is zero.