Video: Using Properties of Limits and Direct Substitution to Find the Limit of a Function at a Point

Given that lim_(π‘₯ β†’ 3) 𝑓(π‘₯)/4π‘₯Β² = βˆ’4, determine lim_(π‘₯ β†’ 3) 𝑓(π‘₯)/π‘₯.

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Video Transcript

Given that the limit as π‘₯ approaches three of 𝑓 of π‘₯ over four π‘₯ squared is equal to negative four, determine the limit as π‘₯ approaches three of 𝑓 of π‘₯ over π‘₯.

To establish this limit, we’re going to recall a few of our limit laws. The first law we might recall is that the limit of the product of two functions is equal to the product of the limit of those two functions. So, what we’re going to begin by doing is writing 𝑓 of π‘₯ over four π‘₯ squared as the product of two functions, as the product of 𝑓 of π‘₯ over π‘₯ and one over four π‘₯.

And the reason we did this is we now have a function that looks a little bit like the limit we’re trying to find. And we can now split this limit up as the limit as π‘₯ approaches three of 𝑓 of π‘₯ over π‘₯ times the limit as π‘₯ approaches three of one over four π‘₯.

We’re now going to apply direct substitution to actually evaluate the limit as π‘₯ approaches three of one over four π‘₯. It’s one over four times three, which is one twelfth. We’re going to put this constant in front of our other limit and we see that our limit becomes a twelfth times the limit as π‘₯ approaches three of 𝑓 of π‘₯ over π‘₯. And of course, this is still our original limit so it’s equal to negative four.

We can now solve this limit equation by multiplying both sides by 12. 12 times negative four is negative 48. And we see that we actually have the solution to this question. The limit as π‘₯ approaches three of 𝑓 of π‘₯ over π‘₯ is equal to negative 48.

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