Evaluate the following limits. Use l’Hôpital’s Rule when it is...

Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→∞ (x - √(x²+4x))

Accepted Answer

Below there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the limit. Use Lahapato's rule if necessary. The limit as X approaches infinity of the square root of X2 + 6 X minus X. Awesome. So it appears for this particular problem, what we're ultimately trying to solve for is we're trying to evaluate this specific limit, and we're given a hint. We're told to use Laha Patal's rule, if necessary, to help us evaluate this specific limit. So with that said, now that we know that we're ultimately trying to figure out how to evaluate this limit and what it's evaluated value will be, let's read off our multiple choice answers to see what our final answer might be. A is -2, B is 0, C is 3, and D is 6. Awesome. So first off, in order to evaluate the limit as X approaches infinity of the square root of X2 + 6 X minus X, we must 1st, 1st off, in order to do that, we need to determine the behavior of the expression. So what does that mean? So we need to note that when we use direct substitution, meaning since we know, or rather we're told that X approaches infinity, so if we were to do direct substitution, if we were to plug in a value of infinity in for X, we need to determine when we do direct substitution, would this You limit yield an indeterminate form. So what does indeterminate what is an indeterminate form? Just in case you don't remem like remember right away what that means. An indeterminate form means you, when you evaluate the limit using direct substitution, that means you get an answer of zero divided by 0, or you get an answer of infinity divided by. Infinity. That's what an indeterminate form is. So with that in mind, we need to use direct substitution in order to determine whether or not our specific limit will yield an indeterminate form, and if it is an indeterminate form, then that means if we get 0 divided by 0 as an answer or infinity divided by infinity, that means we need to use Laha Patal's rule. Awesome. So, for our particular limit, the expression of the indeterminate form, infinity minus infinity, because as X approaches infinity, the, and this is important, so as, let's write this down. So as X approaches infinity, we need to note that the square root of X2 + 6 X. It's going to approach infinity, so we can use algebraic manipulation to evaluate this particular limit. So we need to factor out the X squared under our square root. So when we write this, we need to take the square root of X2 + 6 X. is going to be equal to the square root of X 2 multiplied by 1 + 6 divided by X minus X, fantastic. And we needed to note that since X is greater than 0, that would mean that the square root of X2 is equal to X. Awesome. So the square root of X2 is equal to x. So that means we can go ahead and write that the square root of X2 multiplied by 1 + 6 divided by X. minus X is equal to X, multiplied by the square root of 1 + 6 divided by X minus 1. Fantastic. So thus we can rewrite our limit as the limit as X approaches infinity of the square root of X2 + 6 X minus X is equal to the limit as X approaches infinity of X multiplied by the square root of 1 + 6 divided by X minus 1. Fantastic. So, our next step now is we need to note that as X approaches infinity, that will mean as a result, so once again, as X approaches infinity. The square root of 1 + 6 divided by X minus 1 will approach 0. So this is of the form infinity multiplied by 0. So this is infinity multiplied by 0. So rewriting our limit will give us the following. So if you take the limit as X approaches infinity of X multiplied by the square root of 1 + 6 divided by X minus 1. is equal to the limit, as X approaches infinity of the square root of 1 + 6 divided by X minus 1 divided by 1 divided by X. Awesome. So now we need to note That the new limit, well, when we evaluate it using the direct substitution method that I mentioned before, will yield an indeterminate form which is 0 divided by 0, since both the numerator and the denominator will approach 0 as X approaches infinity. So we need to change the, so we need to use as a result, we need to use the change of variables. So we need to state that T is equal to 1 divided by X, and we need to note that this will be as X approaches infinity and T approaches, so we now need to note that this is as X approaches infinity and as T approaches 0 from the right. So with that in mind, we need to go ahead and write. That as the limit as X approaches infinity of the square root of 1 + 6 divided by X minus 1 divided by 1 divided by X, it's going to be equal to the limit as X approaches 0 from the right. That's what that positive symbol and the power means. That means we're approaching 0 from the right. Oh, so when you take the limit as X approaches 0 from the right of the square root of 1 + 6 divided by X minus 1 divided by T. So now we can recall and apply La Hapital's rule. So first we need to find the derivatives of the numerator and denominator. So focusing our attention on the numerator first, which I denoted as N, so we need to find the derivative of the numerator, so we're trying to find the derivative of the square root of 1 + 6 T minus 1. And this is going to be D, so this derivative is going to be equal to D by DT so we're taking the derivative with respect to T. Of the square root of 1 + 60 minus 1. Fantastic. And this is gonna be equal to 6 divided by 2 multiplied by the square root of 1 + 6 T. And when we simplify, this will be equal to 3 divided by the square root of 1 + 6 T. Fantastic. And we need to note that the derivative of the denominator, which I denoted as D, so we're taking the derivative of T, this will be simply 1 is a derivative of with respect to T of T is gonna just be 1. So now, finally, we can apply Lahapato's rules, we need to take the limit as X approaches 0 from the right of 3. Divided by the square root of 1 + 6T. And we need to note that as, and let's make a note of this up above, we need to note that as T approaches 0 from the right, we need to note that the square root of 1 + 6 T. is approaching the square root of 1, and the square root of 1 is 1. So therefore, without a doubt, we will be able to state that our limit says it's gonna be 3 divided by 1 and 3 divided by 1 is simply equal to 3, and that is our value for the evaluated limit. Hooray, we did it. So looking at our multiple choice answers, the correct answer, without a doubt has to be multiple choice answer letter C, which once again is 3. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.
lim_x→∞ (x - √(x²+4x))

Key Concepts

Limits

l'Hôpital's Rule

Square Root Functions

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