Video: Evaluating Limits Using Limit Laws

Find an exact expression for lim_(π‘₯ β†’ 4) (π‘₯Β³ βˆ’ π‘₯ + 2π‘₯ + 7)/√(π‘₯ + 1) using the limit laws.

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Video Transcript

Find an exact expression for the limit as π‘₯ approaches four of π‘₯ cubed minus π‘₯ plus two π‘₯ plus seven all over the square root of π‘₯ plus one using the limit laws.

The question tells us to find an exact expression for our limit using our limit laws. And this what β€œexact” implies to us that there’s probably going to be an irrational term somewhere. So let’s apply some of our limit laws. The first law we recall is that the limit of the quotient of two functions is equal to the quotient of the limit of each of those function. That’s providing the limit of the denominator is not equal to zero. Well, we know that as π‘₯ approaches four, our denominator does not equal zero. And this means we can write our limit as the limit as π‘₯ approaches four of π‘₯ cubed minus π‘₯ plus two π‘₯ plus seven all over the limit as π‘₯ approaches four of the square root of π‘₯ plus one.

Next, we recall that the limit of the sum or difference of two or more functions is equal to the sum or difference of the limits of those respective functions. And so we rewrite our numerator as shown. And if we also recall that the limit as π‘₯ approaches π‘Ž of the square root of 𝑓 of π‘₯ is equal to the square root of the limit as π‘₯ approaches π‘Ž of that function. Then we can also rewrite our denominator as the square root of the limit as π‘₯ approaches four of π‘₯ plus one.

And then, if we recall that the limit of a constant times a function is equal to the constant times the limit of the function, we see that we can rewrite the limit as π‘₯ approaches four of two of π‘₯ as two times the limit as π‘₯ approaches four of π‘₯. And using the sum law β€” that’s the second law we applied β€” we can also write our denominator as the square root of the limit as π‘₯ approaches four of π‘₯ plus the limit as π‘₯ approaches four of one.

Now, we know that the limit of a constant is just equal to that constant itself. So the limit as π‘₯ approaches four of seven is seven. And the limit as π‘₯ approaches four of one is one. We can also apply direct substitution to our remaining limits. That gives us four cubed minus four plus two times four plus seven all over the square root of four plus one, which simplifies really nicely to 75 over the square root of five.

We have one more step. We’re going to rationalise the denominator of this fraction. To do this, we’re going to multiply both the numerator and the denominator by the square root of five. This has the effect of making the denominator of our fraction a rational number. But also since we’re multiplying by the square root of five over the square root of five, which is the equivalent of one, we don’t actually change the size of our fraction. This becomes 75 root five over five. And since 75 divided by five is 15.

And so we find the limit as π‘₯ approaches four of π‘₯ cubed minus π‘₯ plus two π‘₯ plus seven all over the square root of π‘₯ plus one, using the limit laws, to be 15 root five.

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