Like

Report

The first appearance in print of l'Hospital's Rule was in the book Analyse des Infiniment Petits published by the Marquis de l'Hospital in 1696. This was the first calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to find the limit of the function

$$ y = \frac{\sqrt{2a^3 x - x^4} - a\sqrt[3]{aax}}{a - \sqrt[4]{ax^3}} $$

as $ x $ approaches $ a $, where $ a > 0 $. (At that time it was common to write $ aa $ instead of $ a^2 $.) Solve this problem.

We see that both numerator and denominator approach $0,$ so we can use I'Hospital's Rule:

\[

\begin{aligned}

\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}} & \underline{H} \lim _{x \rightarrow a} \frac{\frac{1}{2}\left(2 a^{3} x-x^{4}\right)^{-1 / 2}\left(2 a^{3}-4 x^{3}\right)-a\left(\frac{1}{3}\right)(a a x)^{-2 / 3} a^{2}}{-\frac{1}{4}\left(a x^{3}\right)^{-3 / 4}\left(3 a x^{2}\right)} \\

&=\frac{\frac{1}{2}\left(2 a^{3} a-a^{4}\right)^{-1 / 2}\left(2 a^{3}-4 a^{3}\right)-\frac{1}{3} a^{3}\left(a^{2} a\right)^{-2 / 3}}{-\frac{1}{4}\left(a a^{3}\right)^{-3 / 4}\left(3 a a^{2}\right)} \\

&=\frac{\left(a^{4}\right)^{-1 / 2}\left(-a^{3}\right)-\frac{1}{3} a^{3}\left(a^{3}\right)^{-2 / 3}}{-\frac{3}{4} a^{3}\left(a^{4}\right)-3 / 4}=\frac{-a-\frac{1}{3} a}{-\frac{3}{4}}=\frac{4}{3}\left(\frac{4}{3} a\right)=\frac{16}{9} a

\end{aligned}

\]

You must be signed in to discuss.

Missouri State University

Campbell University

Oregon State University

University of Nottingham

So for this problem, we are going to be discussing limits. Um, and were given the limit as X approaches A of, um we're just gonna plug in a immediately. So in the numerator we get on the square root of two a cube X minus X to the fourth, and then that's gonna be minus eight times the cube root of a X. So when we plug in a what we're gonna end up getting is that that's equal to the square root of to A to the fourth minus a to the fourth minus a cube times this square or the cube drew of a cube. So it's just eight of the fourth. And then similarly, we will end up with, um eight of the fourth. Here is Well, um, so as a result of that, we know that this is actually a squared minus a squared, which is zero. And then the denominator will have the limit as acts of purchase A of a minus, the the root of X cubed when we do that one of getting a minus thebe route of a cute or a to the fourth. So regardless of what we get here, we know that in plugging in a we're going to get zero as well. So based on this, what that tells us is that we are dealing with the indeterminant form 0/0. So we're gonna wanna use Low Patel's rule. When we differentiate the numerator and denominator, we're gonna end up with the limit as X approaches a, um are numerator is going to be when we have we take the derivative and then we evaluated at a, um we're gonna get the limit. Actually, we can already plug in the values, so let's get rid of this limit. So once we take the derivative, we can just plug in the values, so we're gonna get one half eight of the fourth to the negative one half time to negative to a cube minus a cubed over three times a cube to the negative, two thirds over negative 3/4 a cube age of the fourth to the negative three force and we can do the algebra here. A lot of simplification. Ultimately, what we're going to get down to is 12 a plus for a over nine, which is equal to 16 a over nine, and that will be our final answer

California Baptist University