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Evaluate the limit as π₯, π¦ approaches three, two of π₯ squared π¦ cubed minus four π¦ squared, if it exists.
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In this question, weβre looking to find the limit of a multivariable function, a function which involves more than one variable.
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Here, thatβs π₯ and y.
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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.
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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.
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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?
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Well, the function π₯ squared π¦ cubed minus four π¦ squared is a multivariable polynomial.
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Now, we know that polynomials themselves are continuous over their entire domain.
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The same is true for multivariable polynomials.
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So the function π₯ squared π¦ cubed minus four π¦ squared is continuous over its entire domain.
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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.
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This becomes three squared times two cubed minus four times two squared.
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This simplifies somewhat to nine times eight minus four times four, which is 72 minus 16.
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That simplifies to 56.
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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.