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

Evaluate lim_((π‘₯, 𝑦) β†’ (0, 0)) (π‘₯Β³ βˆ’ 𝑦³)/(π‘₯Β² + π‘₯𝑦 + 𝑦²), if it exists.

02:23

Video Transcript

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

Here, we’re dealing with a multivariable function. It’s a function that involves more than one variable; this is π‘₯ and 𝑦. And so, we recall that if a function in π‘₯ and 𝑦 is continuous at some point π‘Ž, 𝑏, then the limit as π‘₯, 𝑦 approaches that point of 𝑓 of π‘₯, 𝑦 is equal to 𝑓 of π‘Ž, 𝑏. 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 𝑦 cubed over π‘₯ squared plus π‘₯𝑦 plus 𝑦 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 𝑦 equals zero into this function, we get zero over zero, which is of indeterminate form. But let’s see if we can manipulate our function somewhat. Well, the numerator, π‘₯ cubed 𝑦 cubed, is the difference of two cubes. And we know we can write this as π‘₯ minus 𝑦 times π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared. And then, our fraction becomes π‘₯ minus 𝑦 times π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared all over π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared. And we see we have a common factor of π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared. So, we divide through. And our entire fraction simplifies really nicely. And we’re simply left with π‘₯ minus 𝑦.

And so, the question we’re now going to ask ourselves is, is the function π‘₯ minus 𝑦 continuous at zero, zero? π‘₯ minus 𝑦 is a multivariable polynomial. And just like single-variable polynomials, these are continuous over their entire domain. And this means the function π‘₯ minus 𝑦 is continuous over its entire domain, and therefore it must be continuous at zero, zero. And so, to find the limit as π‘₯, 𝑦 approaches zero, zero of π‘₯ cubed minus 𝑦 cubed over π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared, we’re going to find the limit as π‘₯, 𝑦 approaches zero, zero of π‘₯ minus 𝑦. And we’re going to achieve this by letting π‘₯ be equal to zero and 𝑦 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‘₯, 𝑦 approaches zero, zero of π‘₯ cubed minus 𝑦 cubed over π‘₯ squared plus π‘₯𝑦 plus 𝑦 squared is zero.

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