# Question Video: Finding the Limit of a Product of Two Functions Mathematics • 12th Grade

Assuming that lim_(π₯ β 6) β(π₯) = 3, find lim_(π₯ β 6) π₯ β β(π₯).

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### Video Transcript

Assuming that the limit as π₯ approaches six of β of π₯ is equal to three, find the limit as π₯ approaches six of π₯ times β of π₯.

The question tells us that the limit of some function β of π₯ as π₯ approaches six is equal to three. We need to use this information to find the limit as π₯ approaches six of π₯ times the function β of π₯. We can see that π₯ is approaching six in both of these limits, and in our second limit, we have β of π₯. So, we could answer this question directly if we could rewrite our second limit to be in terms of the limit of β of π₯.

And we already know a rule for rewriting the limit of a product of two functions. We know, for any functions π and π and for any real constant π, the limit as π₯ approaches π of π of π₯ times π of π₯ is equal to the limit as π₯ approaches π of π of π₯ times the limit as π₯ approaches π of π of π₯. In other words, the limit of our product is equal to the product of the limits.

We want to apply this to the limit given to us in the question. So, weβll set our function π of π₯ equal to π₯, our function π of π₯ to be β of π₯, and π equal to six. So, by using the fact that a limit of a product is equal to the product of a limit. This gives us the limit as π₯ approaches six of π₯ times β of π₯ is equal to the limit as π₯ approaches six of π₯ times the limit as π₯ approaches six of β of π₯.

We can evaluate our first limit, the limit of π₯ as π₯ approaches six. π₯ is a polynomial, so we can evaluate this by direct substitution. We substitute π₯ is equal to six. This gives us the limit as π₯ approaches six of π₯ is just equal to six. And remember, the question tells us the limit as π₯ approaches six of β of π₯ is equal to three. So, we can evaluate the second limit. Itβs just equal to three. So, our second limit is equal to three. This gives us six times three, which is, of course, equal to 18.

So, weβve shown if the limit as π₯ approaches six of some function β of π₯ is equal to three, then the limit as π₯ approaches six of π₯ times β of π₯ must be equal to 18.