### Video Transcript

Evaluate the limit as π₯, π¦
approaches three, two of π₯ squared π¦ cubed minus four π¦ squared, if it
exists.

In this question, weβre looking to
find the limit of a multivariable function, a function which involves more than one
variable. Here, thatβs π₯ and y. And so we recall that, just like
with single-variable functions, if a function is continuous at π, π, then the
limit as π₯, π¦ approaches π, π of π of π₯, π¦ is equal to π of π, π. Essentially, if a function is
continuous at a point, then to take the limit of the function at that point, we just
plug the values into the function.

So the first thing weβre going to
do is ask ourself, is the function in this question β β thatβs the function given by
π₯ squared π¦ cubed minus four π¦ squared β continuous? Well, the function π₯ squared π¦
cubed minus four π¦ squared is a multivariable polynomial. Now, we know that polynomials
themselves are continuous over their entire domain. The same is true for multivariable
polynomials. So the function π₯ squared π¦ cubed
minus four π¦ squared is continuous over its entire domain.

And so to evaluate the limit as π₯,
π¦ approaches three, two of π₯ squared π¦ cubed minus four π¦ squared, weβre simply
going to substitute π₯ equals three and π¦ equals two into the function. This becomes three squared times
two cubed minus four times two squared. This simplifies somewhat to nine
times eight minus four times four, which is 72 minus 16. That simplifies to 56.

And so we see that the limit as π₯,
π¦ approaches three, two of π₯ squared π¦ cubed minus four π¦ squared is equal to
56.