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

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 $x \to -\infty.$

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

\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

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 $x \to -\infty.$

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

\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 $x \to -\infty$

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

- As a first step, divide both the numerator and the denominator by x-to-the-largest-power that is in the denominator.
- 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

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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:

- As a first step, divide both the numerator and the denominator by x-to-the-largest-power that is in the denominator.
- 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.)

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. 🙂