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

Given that lim_(π‘₯ β†’ βˆ’6) (6𝑓(π‘₯) + 4)/(βˆ’π‘₯ + 1) = 4, find lim_(π‘₯ β†’ βˆ’6) 𝑓(π‘₯).

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Video Transcript

Given that the limit as π‘₯ approaches negative six of six 𝑓 of π‘₯ plus four over negative π‘₯ plus one equals four, find the limit as π‘₯ approaches negative six of 𝑓 of π‘₯.

To answer this question, we’re going to need to recall some important limit laws. The first law we’re interested in says that the limit of a quotient of two functions is equal to the quotient of the limits of each of these functions, provided that the limit of the denominator is not equal to zero.

Since we know that negative negative six plus one is not equal to zero, we can write our limit as the limit as π‘₯ approaches negative six of six times 𝑓 of π‘₯ plus four all over the limit as π‘₯ approaches negative six of negative π‘₯ plus one. And in fact, we can then use direct substitution to work out the limit of our denominator.

It’s negative negative six plus one, which is, of course, seven. And so, we can now rewrite our entire limit. It’s the limit as π‘₯ approaches negative six of six times 𝑓 of π‘₯ plus four all over seven. Now, going back to our original question, we know that this is equal to four. We’re going to make this look a little bit nicer by multiplying both sides of this equation by seven. And when we do, we see that the limit as π‘₯ approaches negative six of six times 𝑓 of π‘₯ plus four is equal to 28.

We’re going to recall a second law of limits. This time, the limit of the sum of two functions is equal to the sum of the limits. So we can say that the limit of six times 𝑓 of π‘₯ plus the limit of four gives us 28. But four itself is independent of π‘₯. And so, we can say that the limit as π‘₯ approaches negative six β€” and in fact, any number β€” of four is simply four. Similarly, in our first limit, six is independent of π‘₯.

And in general, we say that the limit of a constant times a function is equal to the constant times the limit of that function. So we can write the limit as π‘₯ approaches negative six of six times 𝑓 of π‘₯ as six times the limit of 𝑓 of π‘₯. And of course, the sum of these two expressions is equal to 28. So we subtract four from both sides of our equation. And we see that six times the limit as π‘₯ approaches negative six of 𝑓 of π‘₯ is equal to 24.

Our final step to find the limit as π‘₯ approaches negative six of 𝑓 of π‘₯ is to divide both sides of this limit equation by six. And when we do, we find that the limit as π‘₯ approaches negative six of 𝑓 of π‘₯ is equal to four.

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