Video: Evaluating the Limits of a Multivariable Function Using Direct Substitution

Evaluate lim_((π‘₯, 𝑦) β†’ (3, 2)) (π‘₯²𝑦³ βˆ’ 4𝑦²), if it exists.

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Video Transcript

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

In this question, we’re looking to find the limit of a multivariable function, a function which involves more than one variable. Here, that’s π‘₯ and y. And so we recall that, just like with single-variable functions, if a function is continuous at π‘Ž, 𝑏, then the limit as π‘₯, 𝑦 approaches π‘Ž, 𝑏 of 𝑓 of π‘₯, 𝑦 is equal to 𝑓 of π‘Ž, 𝑏. 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 we’re going to do is ask ourself, is the function in this question ⁠— that’s the function given by π‘₯ squared 𝑦 cubed minus four 𝑦 squared β€” continuous? Well, the function π‘₯ squared 𝑦 cubed minus four 𝑦 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 𝑦 cubed minus four 𝑦 squared is continuous over its entire domain.

And so to evaluate the limit as π‘₯, 𝑦 approaches three, two of π‘₯ squared 𝑦 cubed minus four 𝑦 squared, we’re simply going to substitute π‘₯ equals three and 𝑦 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‘₯, 𝑦 approaches three, two of π‘₯ squared 𝑦 cubed minus four 𝑦 squared is equal to 56.

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