Consider the lim_x→∞ (√ ax + b) / √cx + d where a, b, c, ...

Question

Consider the lim_x→∞ (√ ax + b) / √cx + d where a, b, c, and d are positive real numbers. Show that l’Hôpital’s Rule fails for this limit. Find the limit using another method.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the limit using Lopital's rule. Use another method if Lopital's rule is not applicable, and we're given the limit as x approaches infinity of the square root of the quantity of 6 X + 1 in quantity, divided by the square root of the quantity of 2 X + 3 in quantity. We're given 4 possible choices as our answers. For choice A, we have the square root of 2. For choice B, we have the square root of the quantity of 3 divided by 2. For choice C, we have the square root of 3, and for choice D, we have the square root of 6. Now we're asked to evaluate this limit using Lopital's rule, and so we can first try to evaluate our limit here by direct substitution, and in doing so, this is going to be equal to infinity divided by infinity. So here we do have an indeterminant form, so we will be able to apply Lopital's rule to this limit. So we call for Lopital's rule that we need to take the derivative of both our numerator and denominator with respect to X, and then re-evaluate the limit. We'll have the limit as X approaches infinity. Of the derivative with respect to X of the square root. Of the quantity of 6 X plus 1 in quantity, divided by the derivative with respect to X of the square root of the quantity of 2 X plus 3 in quantity. So taking the derivatives, this is going to be equal to the limit. As X approaches infinity. Of quantity, and in the numerator we'll have the derivative with respect to X of the square root of the quantity of 6 X plus 1. So recall that when you take the derivative of a square root, you'll get 1 divided by. 2 multiplied by the original square root, which is going to be the square root of the quantity of 6 X plus 1 in quantity. But we'll need to multiply this by the derivative of what was underneath the radical. So here we're applying our chain rule. So we'll have the derivative with respect to x of the quantity of 6 X plus 1. And so that derivative is equal to 6, so we'll multiply our fraction here by 6. Then this gets divided by the derivative with respect to x of the square root of the quantity of 2 X plus 3. So again, we're taking the derivative of a square root function, so we'll get the quantity of 1 divided by the quantity of 2 multiplied by the square root of the quantity of 2 X plus 3 in quantity. Then we have to multiply this by the derivative with respect to x of the quantity of 2 X plus 3, which is 2. So simplifying this expression, this is going to be equal to. The limit as X approaches infinity. Of 3 multiplied by the square root of the quantity of 2 X plus 3 in quantity, divided by the square root of the quantity of 6 X plus 1 in quantity. And if we try to evaluate this limit by direct substitution, we again, get this to be equal to infinity, divided by infinity. So, we still have an indeterminate form, so let's apply Lopital's rule one more time. So that this is going to be equal to the limit. As X approaches infinity. Of the derivative with respect to x of the quantity of 3 multiplied by the square root of the quantity of 2 X plus 3 in quantity divided by the derivative with respect to x of the square root of the quantity of 6 X plus 1. So again, evaluating these derivatives, this is going to be equal to the limit. As X approaches infinity. Of the quantity and the numerator will have 3 multiplied by the derivative with respect to X of the square root of the quantity of 2 X plus 3. We had found that expression earlier, and that was going to be 1 divided by. 2 multiplied by the square root. Of the quantity of 2 X plus 3 in quantity multiplied by 2. This is divided by the derivative with respect to X of the square root of the quantity of 6 X plus 1. We also found this earlier. And that was the quantity of 1 divided by the quantity of 2 multiplied by the square root of the quantity of 6 X plus 1 in quantity, multiplied by 6. So, simplifying this expression, this is going to be equal to the limit, as X approaches infinity. Of the quantity. Of the square root of 6 X plus 1 in quantity divided by the square root of the quantity of 2 X plus 3. And here we get back our original limit. And so if we continue to apply Lopiel's rule, we're just going to oscillate between the two results that we've already calculated. Our original limit and the limit as x approaches infinity of 3 multiplied by the square root of the quantity of 2 X plus 3 in quantity divided by the square root of the quantity of 6 x + + 1. So, what we need to do is actually use a different method to evaluate this limit. And so we're going to evaluate this limit by factoring out a square root of X out of both the numerator and denominator. So we'll have the limit. As X approaches infinity. Of the square root of X multiplied by the square root. Of 6 Plus 1 divided by X in quantity, and that gets divided by the square root of X, multiplied by the square root of the quantity of 2, plus 3 divided by X in quantity. So the square root of X out front cancels out. And with the remaining um terms, we can actually just evaluate your limit by direct substitution. So this is going to be equal to the square root of the quantity of 6 plus 1 divided by infinity in quantity, divided by the square root of 2, plus 3 divided by infinity. Simplifying this expression, this is going to be equal to the square root of 6 + 0. And that quantity is divided by the square root of the quantity of 2 + 0. In simplifying this expression, this is going to be equal to the square root of 3. So this here is the answer that we are actually looking for. And we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Consider the lim_x→∞ (√ ax + b) / √cx + d where a, b, c, and d are positive real numbers. Show that l’Hôpital’s Rule fails for this limit. Find the limit using another method.

Key Concepts

L'Hôpital's Rule

Limits at Infinity

Dominant Terms

Watch next