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C.8 Limit 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.

Note: Not all courses require you to be able to find limits at $\pm \infty$ of functions that have square roots in them. Please check with your instructor before devoting time to this topic.

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 $x \to \pm \infty$ 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 $x \to +\infty.$ We’ll then take a quick detour to discuss one tricky issue that arises when we wish to consider $x \to -\infty.$

Example 1: $\displaystyle{\lim_{x \to \infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x}}$

Find $\displaystyle{\lim_{x \to \infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x}}.$

Solution.
As we’ve done throughout this Section when working to formally find the limit as $x \to \infty,$ we begin by dividing the numerator and denominator both by x-to-the-largest-power in the denominator, which here is x. Note that to move the x in the numerator underneath the square root sign, we use the fact that $x = \sqrt{x^2}.$
\begin{align*}
\lim_{x \to \infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x} &= \lim_{x \to \infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x} \cdot \dfrac{1/x}{1/x} \\[8px] &= \lim_{x \to \infty}\dfrac{\dfrac{\sqrt{9x^2 + x + 4}}{\sqrt{x^2}}}{\dfrac{2x}{x}} \\[8px] &= \lim_{x \to \infty}\dfrac{\sqrt{\dfrac{9x^2 + x + 4}{x^2}}}{2} \\[8px] &= \lim_{x \to \infty}\dfrac{\sqrt{9 + \dfrac{1}{x} + \dfrac{1}{x^2}}}{2} \\[8px] &= \dfrac{\sqrt{\displaystyle{\lim_{x \to \infty}9} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{1}{x}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{1}{x^2}}}}}{\displaystyle{\lim_{x \to \infty}2}} \\[8px] &= \dfrac{\sqrt{9}}{2} = \dfrac{3}{2} \quad \cmark
\end{align*}
You can use the interactive calculator below to see how the function’s curve mimics the horizontal asymptote $y = \dfrac{3}{2}$ as x grows large. (Actually, you’ll note that this particular function approaches its horizontal asymptote quite rapidly.)

Graph of $f(x) = \dfrac{\sqrt{9x^2 + x + 4}}{2x}$ versus x to examine behavior as $x \to \infty$

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 $x = \sqrt{x^2}$ under the square-root sign and the various manipulations take a bit of practice.

Quick discussion of a subtle and important issue as $x \to -\infty$
But before then, let’s address the one tricky issue that arises when we look at $x \to -\infty.$ 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 here.) If we proceed without much thought, we would once again make our substitution $x = \sqrt{x^2}$ . . . and substituting that value $x = -5,$ we have $-5 = \sqrt{25} = 5.$ But wait, that’s not right: $5 \ne -5!$

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:
\[\text{Absolute value definition: }\quad \lvert x\rvert = \sqrt{x^2} \] “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 $x = \sqrt{x^2}$ 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:”
\[ \bbox[10px,border:2px solid blue]{\text{For negative }x: \quad x = -\sqrt{x^2}}\] One way to frame this whole issue is that if you have the equation $x^2 = 25,$ then you know there are two possible answers, one positive and one negative: $x = \bbox[yellow]{\pm} \sqrt{25} = \bbox[yellow]{\pm} 5.$ 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 $x \to -\infty.$

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 $x \to -\infty.$

Example 2: $\displaystyle{\lim_{x \to -\infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x}}$

Find $\displaystyle{\lim_{x \to -\infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x}}.$

Solution.
Before we do anything else, let’s look at the function and decide whether we expect the limit — if it exists (as it typically does in these problems) — will be positive or negative. We can reason quickly: looking at $\dfrac{\sqrt{9x^2 + x + 4}}{2x},$ the numerator is always positive because of the (positive) 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 $x \to -\infty,$ 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.

As usual, we first divide the numerator and denominator both by x-to-the-largest-power in the denominator, which here is x. Note that to move the x in the numerator underneath the square root sign, we use the fact that since we’re concerned only with negative values of x, we have $x = \bbox[yellow]{-}\sqrt{x^2},$ as discussed immediately above this Example.
\begin{align*}
\lim_{x \to -\infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x} &= \lim_{x \to -\infty}\dfrac{\sqrt{9x^2 + x + 4}}{2x} \cdot
\dfrac{1/x}{1/x} \\[8px] &= \lim_{x \to -\infty}\dfrac{\dfrac{\sqrt{9x^2 + x + 4}}{\left( \bbox[yellow]{-}\sqrt{x^2}\right)}}{\dfrac{2x}{x}} \\[8px] &= \lim_{x \to -\infty}\dfrac{\bbox[yellow]{-}\sqrt{\dfrac{9x^2 + x + 4}{x^2}}}{2} \\[8px] &= \lim_{x \to -\infty}\dfrac{\bbox[yellow]{-}\sqrt{9 + \dfrac{1}{x} + \dfrac{1}{x^2}}}{2} \\[8px] &= \dfrac{\bbox[yellow]{-}\sqrt{\displaystyle{\lim_{x \to -\infty}9} + \cancelto{0}{\displaystyle{\lim_{x \to -\infty}\dfrac{1}{x}}} +
\cancelto{0}{\displaystyle{\lim_{x \to -\infty}\dfrac{1}{x^2}}}}}{\displaystyle{\lim_{x \to -\infty}2}} \\[8px] &= \bbox[yellow]{-}\dfrac{\sqrt{9}}{2} = \bbox[yellow]{-}\dfrac{3}{2} \quad \cmark
\end{align*}
We highlighted the negative sign in every step after we introduced it so you can see how it directly gives us the negative result we expect and that is correct: the limit as $x \to -\infty$ is $-\dfrac{3}{2}.$ If we didn’t insert the negative sign $x = -\sqrt{x^2}$ “by hand” in line #2 of our solution, we would obtain a positive result in the end, which is clearly incorrect.

Graph of $f(x) = \dfrac{\sqrt{9x^2 + x + 4}}{2x}$ versus x to examine behavior as $x \to -\infty$

Tips icon to highlight tricky point about finding the limit at infinity of a function with a square rootLet’s summarize the approaches we used in Examples 1 and 2:

  1. As a first step, divide both the numerator and the denominator by x-to-the-largest-power that is in the denominator.
  2. You already know $x = \pm \sqrt{x^2}.$ You must now remember to choose which sign is appropriate in the given situation:
    \begin{align*}
    \text{For }x \to \infty: \quad x &= \sqrt{x^2} \\[8px] \text{For }x \to -\infty: \quad x &= -\sqrt{x^2}
    \end{align*}
    In particular, this means that inthe case of $x \to -\infty,$ you must insert the negative sign “by hand.”


As you’ll see in the problems below, there is one additional manipulation we’ve used before that proves necessary here as well: multiplying by the conjugate of the function divided-by-itself. We strongly suggest that you work through the problems below for yourself to integrate the techniques into your repertoire.

Practice Problems: Limits at Infinity with Square Roots

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The Upshot

This is a screen about the technique for how to find the limit in a very particular case, when you’re computing the limit at infinity of a function with a square root. Repeating the key new problem-solving steps from above:

  1. As a first step, divide both the numerator and the denominator by x-to-the-largest-power that is in the denominator.
  2. You already know $x = \pm \sqrt{x^2}.$ You must now remember to choose which sign is appropriate in the given situation:
    \begin{align*}
    \text{For }x \to \infty: \quad x &= \sqrt{x^2} \\[8px] \text{For }x \to -\infty: \quad x &= -\sqrt{x^2}
    \end{align*}
    In particular, this means that in the case of $x \to -\infty,$ you must insert the negative sign “by hand.”


These problems can be tricky! If you need help, please pop over to the Forum and ask. (The easy-to-use math editor there will even make it simple to recreate your question with nice equation rendering for us all to see and be able to understand what can be complicated-looking functions.)


[What’s below repeats from the preceding screen, in case you had to delay finishing this Section until after including this screen.]
This concludes our exploration of limits at $\pm \infty.$ You now have many tools to reason about a function’s behavior as $x \to -\infty$ and $x \to \infty!$

In the next section, we’ll take up the important concept of “continuity” and “continuous functions,” along with the related “Intermediate Value Theorem.” We’ll see you there. 🙂