Limits of quotients

Find the...

Question

Limits of quotients

Find the limits in Exercises 23–42.

limx→−3 (2 − √(x² − 5)) / (x + 3)

Accepted Answer

Welcome back everyone. In this problem, we want to determine the limit as X approaches -6 of 5 minus the square root of X2 minus 11 divided by X + 6. A says it's -65, B 6/5, C12/5, and the D-125. Now, to determine this limit, let's first try by direct substitution. If we directly substitute X equals -6, OK. Into our equation. The note is here that our denominator is going to be -6 + 6, which equals 0. And if our denominator is 0, then it leads to an indeterminate form. So by direct substitution. OK, then. Or rather we can say, let me, let me rewrite this, OK? Di direct substitution. Leads to an indeterminate form. And since that's the case, that means we'll need to evaluate our limit in another way so that it does not end up in an indeterminate form. So how can we do that? Well, we could multiply our numerator and denominator by the conjugate. Then we can simplify and see if after we simplify and substitute the value of X directly, if it will lead to an indeterminate form, OK. So let's try to do that. Let's multiply, multiplying by the conjugate. It's multiplied by the conjugate. And now when we do that, OK. Then we're going to have the limit as X approaches -6 of 5 minus the square root of X2 minus 11 divided by X + 6 multiplied by the conjugate of or numerator, so that's 5 plus the square root of X2 minus 11 divided by 5 plus the square root of X2 minus 11. Now when we do that, let me make some space and come down further here. Then we, it's gonna become the limit as X approaches -6 of 25 minus X2 minus 11. Divided by X + 6. OK. X + 6. Multiplied By 5 plus the square root of X2 minus 11. Now, if we simplify what's going on in our numerator here. OK, this is gonna become 25 minus X2 minus -11, that's 25 + 11. And thus we're gonna have the limit then as X approaches -6 of 25 + 11 or 36 minus X2 divided by X + 6. Multiplied by 5 plus the square root of X2 minus 11. Now notice that our numerator represents the difference of two squares. So then we can rewrite this as X + 6, sorry. 6 Plus X, which is the same as X + 6, but I'm just writing it for clarity here to see where it comes from, multiplied by 6 minus X. Divided by X + 6 multiplied by 5 plus the square root of X2 minus 11. Now we can cancel 6 + X and X + 6. Both of them are the same thing. And thus that leaves us, let me make some space on the right here, that now leaves us with the limit as X approaches -6 of 6 minus X. Divided by 5 plus the square root of X2 minus 11. Now, let's evaluate the simplified limit expression as X approaches -6. If we directly substitute no, then it's gonna be equal to 6 minus -6. Divided by 5 plus the square root of -6. Squared minus 11. Now we know here 6 minus -6 is 6 + 6, which is 12. And in our denominator, that's gonna be 5 plus the square of 36 minus 11, or square rate of 25, square of 25 is 5 and 5 + 5 is 10. So this is gonna be 12/10 or simplified 6/5. Thus, our limit is 6/5. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Limits of quotients

Find the limits in Exercises 23–42.

limx→−3 (2 − √(x² − 5)) / (x + 3)

Key Concepts

Limits

Rational Functions

L'Hôpital's Rule

Watch next