Use l’Hôpital’s Rule to find the limits in Exercises 85–108.

Question

95. lim(x→∞) (√(x² + x + 1) - √(x² - x))

Accepted Answer

Hello there. Today we are 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. Find the limit, the limit as x approaches infinity of the square root of x 2 + 3 x + 2 minus the square root of x 2 minus x. OK. So for this problem, we are asked to simply find the limit that is given to us by the prom itself. So now that we know that we're ultimately trying to determine what the limit is, our first step that we need to take is we need to let capital L, which I'll just call L for short, we'll let L equal the limit as X approaches infinity, specifically positive infinity. Of the square root of x to the power of 2. Plus 3 x + 2 minus the square root of x 2 minus x. And we can begin. To solve for our limit, and we can say that L is equal to the limit as X approaches infinity of the square root of, so when we pull out our value. Of x2, so we'll have the square root of x2 multiplied by parentheses of 1 + 3 divided by x + 2 divided by x 2 minus the square root of x 2 multiplied by parentheses of 1 minus 1 divided by x. Fantastic. And we can do a little bit of simplifying, and we can say that L is equal to the limit as X approaches positive infinity of X multiplied by the square root of 1 + 3 divided by X + 2 divided by X minus X multiplied by the square root. Of 1 minus 1 divided by X. Fantastic. So now we need to do is we need to evaluate, so let's make a note here. We need to evaluate 3 divided by X, 2 divided by X to the power of 2, which I forgot. So going back to our L is equal to the limit as X approaches infinity of X multiplied by the square root of 1 + 3 divided by X + 2 divided by X 2. So, it's gonna be 2 divided by X to the power of 2. To be precise, it's always good to make sure when you're solving that you don't forget any powers, so. We need to note, going back to our statement in blue here, we now are evaluating. We need to take the evaluation of 3 divided by X, 2 divided by X to the power of 2. And we need to evaluate 1 divided by X, and we need to note this as it approaches 0. As, once again, this will approach 0 as X approaches positive infinity. Fantastic. So that means drum roll, that therefore, we can go ahead and write that L is equal to square brackets of. Infinity. Minus infinity. Fantastic, which will lead to an indeterminate response. So that means that we will need to use algebraic manipulation to solve. So once again, we can find, we will determine that L is equal to square brackets of infinity minus infinity, which will lead to an indeterminate value, which will require us to use algebraic manipulation in order to solve for this limit. So with that said, we now need to multiply by the conjugate in order to simplify. So we need to note that we could say that l is equal to the limit as x approaches infinity of parentheses of square root of. X2 + 3 x + 2 minus the square root of x 2 minus x multiplied by the square root of x 2 + 3 x + 2. Plus the square root of x 2 minus x all divided by the square root of x 2 + 3 x + 2. Plus the square root of x 2 minus X. Fantastic. So, we will find that L will now be equal to the limit as X approaches infinity of X2 + 3 x + 2 minus x 2. Plus X all divided by the square root of x 2 + 3 x + 2. Plus the square root of x 2 minus X. So, we can go even further and write that L is equal to the limit as X approaches infinity of 4 x + 2. Divided by the square root of x to the power of 2. Plus 3 X. Plus 2, so it's gonna be. Also, so once again, so we'll have 4 x + 2, all divided by the square root of X2 + 3 X + 2. Plus the square root of x 2 minus X. Fantastic. So now, what we need to do, we can say this is our second step, is we can divide the numerator and the denominator by X, and let us make a note that X is equal to the square root of x 2. And when we do that, we will find that L is equal to the limit as X approaches infinity of 4 + 2 divided by X, all divided by the square root of 1 + 3 divided by X plus 2 divided by x to the 2. Plus the square root of 1 minus 1 divided by x. And now, our third step is we need to evaluate, and we are evaluating 2 divided by X, 3 divided by X, 2 divided by X to the power of 2. And -1 divided by x as it approaches 0 as x approaches positive infinity. So therefore, For our fourth and final step we could say that l is equal to square brackets of 4 divided by the square root of 1 plus the square root of 1, which will mean that l is equal to square brackets of 4 divided by 2, which will mean that l is simply equal to 2 and 2 is our final answer. Hooray, we did it. So going back to the top to review the problem itself, we could safely say that our limit, without a doubt, our final answer for the limit once again is 2. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.95. lim(x→∞) (√(x² + x + 1) - √(x² - x))

Watch next

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
95. lim(x→∞) (√(x² + x + 1) - √(x² - x))