C.8 Limits at Infinity with Square Roots

On this screen we look at a very particular type of problem, one where you're finding a limit at infinity with a square root in the function.

There are no new concepts here. There are, however, some useful manipulations you should know about to be able to find these limits.

Let's say upfront: unlike the limits we've examined on the proceeding few screens where dominance was clear, when you're looking at a function that involves square roots it is very difficult to immediately know what the limit is as 𝑥 → ±∞ at first glance. Indeed, the answers are often surprising (as you'll see in the problems below). For this reason, in order to be sure of your answer you must compute the limit using the technique of multiplying by the conjugate that we developed earlier.

Fortunately, those same techniques will prove useful for the problems on this screen as well, as we'll see in the following example that looks at a function's behavior as 𝑥 → +∞. We'll then take a quick detour to discuss one tricky issue that arises when we wish to consider 𝑥 → −∞.

Example 1 is straightforward enough, and we'll let you try some practice problems below to find the limit at infinity with square roots, since moving the 𝑥 =√𝑥2 under the square-root sign and the various manipulations take a bit of practice.

Quick discussion of a subtle and important issue as 𝑥 → −∞

But before then, let's address the one tricky issue that arises when we look at 𝑥 → −∞. First, notice that when we're looking at this limit as x takes on increasingly negative values, x itself is always negative. That's an obvious point, but one that we must keep clearly in mind as we proceed.

Now, to illustrate the issue, let's consider the particular value 𝑥 = −5. (That's obviously not a large number, but it will let us make the relevant point.) If we proceed without much thought, we would once again make our substitution 𝑥 =√𝑥2 . . . and substituting that value 𝑥 = −5, we have −5 =√25 =5. But wait, that's not right: 5 ≠ −5!

What just happened?!? Actually, what we just saw is useful to remember: one way to define the absolute value of a number is to take the (positive) square-root of that number's square: Absolute value definition: |𝑥|=√𝑥2 "Of course," you say: "the square-root always returns a positive number (or zero)."

But in our situation, with 𝑥 = −5, we don't want the absolute value, which is what we accidentally got when we mindlessly set 𝑥 =√𝑥2 a few liines above. Instead, we must retain the negative sign so that 𝑥 = −5 remains a negative number after squaring and taking its square-root. And to achieve that, we must insert the negative sign "by hand":

One way to frame this whole issue is that if you have the equation 𝑥2 =25, then you know there are two possible answers, one positive and one negative: 𝑥 = ±√25 = ±5. You then have to choose which of the plus-and-minus value(s) are the one(s) you need to keep. In this case, we need to keep only the negative choice, 𝑥 = −5, because we know we're dealing with only negative numbers as we look at 𝑥 → −∞.

If this discussion seems abstract, please just continue on to the following example where we'll see how this substitution works in practice as we determine the limit as 𝑥 → −∞.

Let's summarize the approaches we used in Examples 1 and 2:

Of course the only way to really get this down is to practice, so here are some problems to try — each of course with a complete solution.

Practice Problems: Limits at Infinity with Square Roots

Practice Problem 1

(a)

Find lim𝑥→∞√5𝑥2+2𝑥𝑥.

(b)

Find lim𝑥→−∞√5𝑥2+2𝑥𝑥.

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• Solution (a) Detail
• Solution (b) Detail

• (a) √5
• (b) −√5

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We use our usual "trick" of dividing the numerator and the denominator by the largest term in the denominator, which here is 𝑥.

Note that since we're looking at 𝑥 →∞, we're interested only in positive values of 𝑥, and so we'll use the fact that 𝑥 =√𝑥2.

lim𝑥→∞√5𝑥2+2𝑥𝑥=lim𝑥→∞√5𝑥2+2𝑥𝑥⋅1√𝑥21√𝑥2=lim𝑥→∞√5𝑥2+2𝑥√𝑥2𝑥𝑥=lim𝑥→∞√5𝑥2+2𝑥𝑥21=lim𝑥→∞√5+2𝑥=√√ √ √⎷lim𝑥→∞5+lim𝑥→∞(2𝑥)0=√5✓

Note that in the last step, we used the fact that lim𝑥→ ∞2𝑥=0.

We can verify the result with a quick look at the graph of the function. Note that the horizontal line 𝑦 =√5 is a horizontal asymptote for this graph.

Before we do anything else, let's look at the function and decide whether we expect the limit — if it exists (as it typically will in these problems) — will be positive or negative. We can reason quickly: in √𝑥2(5+2𝑥)𝑥, the numerator will always be positive because of the square root. The denominator, on the other hand, will always be negative, because we're looking at ever-increasing negative values for x. Hence as 𝑥 → −∞, the fraction will always have a negative value, and so if we find a number as the limit, we expect it to be negative. This quick initial reasoning is a good check to use against our final result.

To obtain that result, we again use our usual "trick" of dividing the numerator and the denominator by the largest term in the denominator, which here is 𝑥.

The crucial part of this solution: since we're looking at 𝑥 → −∞, we're interested only in negative values of 𝑥, and so we'll use the fact that 𝑥 = −√𝑥2.

lim𝑥→−∞√5𝑥2+2𝑥𝑥=lim𝑥→−∞√5𝑥2+2𝑥−√𝑥2𝑥𝑥=lim𝑥→−∞−√5𝑥2+2𝑥𝑥21=lim𝑥→−∞−√5+2𝑥=−√5✓

Note that in the last step, we used the fact that lim𝑥→−∞2𝑥=0.

We can verify the result with a quick look at the graph of the function. Note that the horizontal line 𝑦 = −√5 is a horizontal asymptote for this graph.

Notice that we obtained a negative number as our answer, which matches our quick initial reasoning above.

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Practice Problem 2

Find lim𝑥→∞(√𝑥2+𝑥−𝑥). We think this problem has a cool, surprising result.

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As 𝑥 grows and Grows, both √𝑥2+𝑥 and 𝑥 grow and Grow too. We thus don't immediately know what the difference between the two terms is. ( ″∞ −∞ ″ could be anything — it is an "indeterminate expression," meaning we have more work to do.) To proceed, we'll use the same approach we used earlier when evaluating limits that had square roots in them: we'll rationalize the expression by multiplying by its conjugate √𝑥2+𝑥 +𝑥 divided by itself: lim𝑥→∞(√𝑥2+𝑥−𝑥)=lim𝑥→∞(√𝑥2+𝑥−𝑥1⋅√𝑥2+𝑥+𝑥√𝑥2+𝑥+𝑥)=lim𝑥→∞(√𝑥2+𝑥)2+𝑥√𝑥2+𝑥−𝑥√𝑥2+𝑥−𝑥2√𝑥2+𝑥+𝑥=lim𝑥→∞(𝑥2+𝑥)−𝑥2√𝑥2+𝑥+𝑥=lim𝑥→∞𝑥√𝑥2+𝑥+𝑥 Let's now use our usual trick of dividing the numerator and the denominator by the largest power in the denominator. That power is 𝑥: while there is an 𝑥2 present, it is under a square root (√𝑥2+...), and so its effective power is 𝑥1. Since we're looking at 𝑥 →∞ we're interested only in positive values of 𝑥, and so we have 𝑥 =√𝑥2. 𝑥2+𝑥=lim𝑥→∞𝑥𝑥√𝑥2+𝑥+𝑥𝑥=lim𝑥→∞1√𝑥2+𝑥𝑥2+1=lim𝑥→∞1√1+1𝑥+1=11+1=12✓ This limit is unexpected, at least to us! But you can check a few numbers to see how it works: 𝑓(𝑥)=√𝑥2+𝑥−𝑥𝑓(10)=√100+10−10≈10.488−10=0.488𝑓(20)=√400+20−20≈20.494−20=0.494𝑓(100)=√10,000+100−100≈100.499−100=0.499

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