Limit at Infinity Problems with Square Roots

Limits at Infinity with Square Roots: Problems and Solutions

To analyze limit at infinity problems with square roots, we’ll use the tools we used earlier to solve limit at infinity problems, PLUS one additional bit: it is crucial to remember

\[ \bbox[yellow,5px]{
\begin{align*}
\text{If $x$ is positive: } x &= \sqrt{x^2} \\[8px] \text{If $x$ is negative: } x &= -\sqrt{x^2} \\[8px] \end{align*} } \]

• For example, if $x = 3$, then $x = 3 = \sqrt{9}$.
• By contrast, if $x = -3$, then $x = -3 = -\sqrt{9}$.

$Tips icon$ You must remember that $x = -\sqrt{x^2}$ in any problem where $x \to\, -\infty$, since you’re then automatically looking at negative values of x.

The problems below illustrate, starting with part (b) of the first one.

Practice Problem #1

(a) Find $\displaystyle{\lim_{x \to \infty}\frac{\sqrt{5x^2 + 2x}}{x}}.$

(b) Find $\displaystyle{\lim_{x \to\, -\infty}\frac{\sqrt{5x^2 + 2x}}{x}}.$

Click to View Calculus Solution

Solution (a)Solution (b)

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 $x \to \infty$, we’re interested only in positive values of $x$, and so we’ll use the fact that $x = \sqrt{x^2}$.
\[ \begin{align*}
\lim_{x \to \infty}\frac{\sqrt{5x^2 + 2x}}{x} &= \lim_{x \to \infty}\frac{\dfrac{\sqrt{5x^2 + 2x}}{\sqrt{x^2}}}{\dfrac{x}{x}} \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{\dfrac{5x^2 + 2x}{x^2}}}{1} \\[8px] &= \lim_{x \to \infty} \sqrt{5 + \dfrac{2}{x}} \\[8px] &= \sqrt{5} \quad \cmark
\end{align*} \]

Note that in the last step, we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{2}{x} = 0}$.

In the limit that x goes to infinity, the curve y=f(x) tends toward y = sqrt(5)
We can verify the result with a quick look at the graph of the function. Note that the horizontal line $y = \sqrt{5}$ 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 $\frac{\sqrt{x^2\left( 5 + \frac{2}{x} \right)}}{x}$, 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 $x \to\, -\infty$, 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.

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 $x \to\, -\infty$, we’re interested only in negative values of $x,$ and so we’ll use the fact that $x = -\sqrt{x^2}.$
\[ \begin{align*}
\lim_{x \to\, -\infty}\frac{\sqrt{5x^2 + 2x}}{x} &= \lim_{x \to\, -\infty}\frac{\dfrac{\sqrt{5x^2 + 2x}}{-\sqrt{x^2}}}{\dfrac{x}{x}} \\[8px] &= \lim_{x \to\, -\infty}\frac{-\sqrt{\dfrac{5x^2 + 2x}{x^2}}}{1} \\[8px] &= \lim_{x \to\, -\infty} -\sqrt{5 + \dfrac{2}{x}} \\[8px] &= -\sqrt{5} \quad \cmark
\end{align*} \] Note that in the last step, we used the fact that $\displaystyle{\lim_{x \to\, -\infty}\frac{2}{x} = 0}$.

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

In the limit that x goes to negative infinity, the curve y = f(x) tends toward y = -sqrt(5) We can verify the result with a quick look at the graph of the function. Note that the horizontal line $y = -\sqrt{5}$ is a horizontal asymptote for this graph.

Practice Problem #2

We think this problem has a cool, surprising result.
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{x^2 + x} – x \right)}.$

Click to View Calculus Solution

As $x$ grows and Grows, both $\sqrt{x^2 + x}$ and $x$ grow and Grow too. We thus don’t immediately know what the difference between the two terms is. $(“\infty – \infty”$ could be anything — it is an “indeterminate expression,” meaning we have more work to do.)

\[ \begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + x} – x \right) &= \lim_{x \to \infty}\left(\frac{\sqrt{x^2 + x} – x}{1} \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}\right) \\[8px] &= \lim_{x \to \infty}\frac{\left( \sqrt{x^2 + x}\right)^2 + x\sqrt{x^2 + x} -x \sqrt{x^2 + x} -x^2 }{\sqrt{x^2 + x} + x} \\[8px] &= \lim_{x \to \infty}\frac{(x^2 + x) -x^2}{\sqrt{x^2 + x} + x} \\[8px] &= \lim_{x \to \infty}\frac{x}{\sqrt{x^2 + x} + x}
\end{align*} \] 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 $x^2$ present, it is under a square root $\left(\sqrt{x^2 + …} \right)$, and so its effective power is $x^1$.

Since we’re looking at $x \to \infty$ we’re interested only in positive values of $x$, and so we have $x = \sqrt{x^2}.$
\[ \begin{align*}
\phantom{x^2 + x} &= \lim_{x \to \infty}\frac{\dfrac{x}{x}}{\dfrac{\sqrt{x^2 + x} + x}{x}} \\[8px] &= \lim_{x \to \infty}\frac{1}{\sqrt{\dfrac{x^2 + x}{x^2}} + 1} \\[8px] &= \lim_{x \to \infty}\frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \\[8px] &= \frac{1}{1+1} \\[8px] &= \frac{1}{2} \quad \cmark
\end{align*} \]

This limit is unexpected, at least to us! But you can check a few numbers to see how it works:
\[ \begin{align*}
f(x) &= \sqrt{x^2 + x} – x \\[8px] f(10) &= \sqrt{100 + 10} – 10 \approx 10.488 – 10 = 0.488 \\[8px] f(20) &= \sqrt{400 + 20} – 20 \approx 20.494 – 20 = 0.494 \\[8px] f(100) &= \sqrt{10,000 + 100} – 100 \approx 100.499 -100 = 0.499
\end{align*} \]

Practice Problem #3

This problem is by student request. It has another (the same, actually) cool, surprising result.
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{x + \sqrt{x}} – \sqrt{x} \right)}$.

Click to View Calculus Solution

As $x$ grows and Grows, both $\sqrt{x + \sqrt{x}}$ and $\sqrt{x}$ grow and Grow too. We thus don’t immediately know what the difference between the two terms is. $(“\infty – \infty”$ could be anything — it is an “indeterminate expression,” meaning we have more work to do.)

\[ \begin{align*}
\lim_{x \to \infty}\left(\sqrt{x + \sqrt{x}} – \sqrt{x} \right) &= \lim_{x \to \infty}\left(\frac{\sqrt{x + \sqrt{x}} – \sqrt{x}}{1} \cdot \frac{\sqrt{x + \sqrt{x}} + \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \right) \\[8px] &= \lim_{x \to \infty}\frac{\left(\sqrt{x + \sqrt{x}} \right)^2 – \sqrt{x}\sqrt{x + \sqrt{x}} + \sqrt{x}\sqrt{x + \sqrt{x}} – \left( \sqrt{x}\right)^2}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty} \frac{\left(x + \sqrt{x} \right) – x}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] \end{align*} \] Let’s now use our usual trick of dividing the numerator and the denominator by the largest power in the denominator. That power is $\sqrt{x}.$
\[ \begin{align*}
&= \lim_{x \to \infty}\frac{\dfrac{\sqrt{x}}{\sqrt{x}}}{\dfrac{\sqrt{x + \sqrt{x}}+ \sqrt{x}}{\sqrt{x}}} \\[8px] &= \lim_{x \to \infty}\frac{1}{\dfrac{\sqrt{x + \sqrt{x}}}{\sqrt{x}}+ \dfrac{\sqrt{x}}{\sqrt{x}}} \\[8px] &= \lim_{x \to \infty}\frac{1}{\sqrt{\dfrac{x + \sqrt{x}}{x}}+ 1} \\[8px] &= \lim_{x \to \infty}\frac{1}{\sqrt{1 + \dfrac{1}{\sqrt{x}}}+ 1} \\[8px] &= \frac{1}{\sqrt{1 + \cancelto{0}{\dfrac{1}{\sqrt{x}}}}+ 1} \\[8px] &= \frac{1}{\sqrt{1}+ 1} \\[8px] &= \dfrac{1}{2} \quad \cmark
\end{align*} \]

Note that toward the end, we used the fact that $\displaystyle{\lim_{x \to\, \infty}\frac{1}{\sqrt{x}} = 0}$.

This limit is unexpected, at least to us! But you can check a few numbers to see how it works:
\[ \begin{align*}
f(x) &= \sqrt{x + \sqrt{x}} – \sqrt{x} \\[8px] f(100) &= \sqrt{100 + \sqrt{100}} – \sqrt{100} \approx 10.48 – 10 = 0.48 \\[8px] f(10,000) &= \sqrt{10,000 + 100} – 100 \approx 100.499 – 100 = 0.499 \\[8px] \end{align*} \]

Practice Problem #4

This is a generalized version of Problem #2 above.
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{a^2 x^2 + x} -ax \right)},$ where $a$ is a positive constant.

Click to View Calculus Solution

As $x$ grows and Grows, both $\sqrt{a^2 x^2 + x}$ and $ax$ grow and Grow too. We thus don’t immediately know what the difference between the two terms is. $(“\infty – \infty”$ could be anything — it is an “indeterminate expression,” meaning we have more work to do.)

\[ \begin{align*}
\lim_{x \to \infty}\left(\sqrt{a^2 x^2 + x} -ax \right) &= \lim_{x \to \infty}\left(\frac{\sqrt{a^2 x^2 + x} -ax}{1}\cdot\frac{\sqrt{a^2 x^2 + x} + ax}{\sqrt{a^2 x^2 + x} + ax} \right) \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{a^2 x^2 + x}\sqrt{a^2 x^2 + x} + ax \sqrt{a^2 x^2 + x} – ax \sqrt{a^2 x^2 + x} – (ax)(ax)}{\sqrt{a^2 x^2 + x} + ax} \\[8px] &= \lim_{x \to \infty}\frac{\left(a^2 x^2 + x \right) -a^2 x^2}{\sqrt{a^2 x^2 + x} + ax} \\[8px] &= \lim_{x \to \infty}\frac{x}{\sqrt{a^2 x^2 + x} + ax}
\end{align*} \]

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 $x^2$ present, it is under a square root $\left(\sqrt{x^2 + …} \right)$, and so its effective power is $x^1.$

Since we’re looking at $x \to \infty$ we’re interested only in positive values of $x$, and so we have $x = \sqrt{x^2}$.

\[ \begin{align*}
\phantom{\sqrt{a^2 x^2 + x} -ax } &= \lim_{x \to \infty}\frac{\dfrac{x}{x}}{\dfrac{\sqrt{a^2 x^2 + x} + ax}{x}} \\[8px] &= \lim_{x \to \infty}\frac{1}{\dfrac{\sqrt{a^2 x^2 + x}}{\sqrt{x^2}} + a} \\[8px] &= \lim_{x \to \infty}\frac{1}{\sqrt{a^2 + \frac{1}{x}} + a} \\[8px] &= \frac{1}{a + a} \\[8px] &= \frac{1}{2a} \quad \cmark
\end{align*} \] Notice in the second-to-last step we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$

Practice Problem #5

Find $\displaystyle{\lim_{x \to \infty} \left(\sqrt{x^2 + ax} – \sqrt{x^2 + bx} \right)},$ where $a$ and $b$ are constants.

Click to View Calculus Solution

As $x$ grows and Grows, both $\sqrt{x^2 + ax}$ and $\sqrt{x^2 + bx}$ grow and Grow too. We thus don’t immediately know what the difference between the two terms is. $(“\infty – \infty”$ 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 $\sqrt{x^2 + ax} + \sqrt{x^2 + bx}$ divided by itself:
\[ \begin{align*}
\lim_{x \to \infty} \left(\sqrt{x^2 + ax} – \sqrt{x^2 + bx} \right) &= \lim_{x \to \infty} \left(\frac{\sqrt{x^2 + ax} – \sqrt{x^2 + bx}}{1} \cdot \frac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \right) \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{x^2 + ax}\sqrt{x^2 + ax} + \sqrt{x^2 + ax}\sqrt{x^2 + bx} – \sqrt{x^2 + bx}\sqrt{x^2 + ax} – \sqrt{x^2 + bx}\sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\[8px] &= \frac{\left(\sqrt{x^2 + ax}\right)^2 – \left(\sqrt{x^2 + bx} \right)^2}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\[8px] &= \lim_{x \to \infty}\frac{\left(x^2 + ax \right) – \left(x^2 + bx \right)}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\[8px] &= \lim_{x \to \infty}\frac{ax – bx}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\[8px] \end{align*} \]

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 $x^2$ present, it is under a square root $\left(\sqrt{x^2 + …} \right)$, and so its effective power is $x^1$.

Since we’re looking at $x \to \infty$ we’re interested only in positive values of $x$, and so we have $x = \sqrt{x^2}.$

\[ \begin{align*}
\phantom{x^2 + ax} &= \lim_{x \to \infty}\frac{\dfrac{ax – bx}{x}}{\dfrac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2}}} \\[8px] &= \lim_{x \to \infty}\frac{a – b}{\sqrt{\dfrac{x^2 + ax}{x^2}} + \sqrt{\dfrac{x^2 + bx}{x^2}}} \\[8px] &= \lim_{x \to \infty}\frac{a-b}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} \\[8px] &= \frac{a-b}{1 + 1} \\[8px] &= \frac{a-b}{2} \quad \cmark
\end{align*} \] Notice in the second-to-last step we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$

Practice Problem #6

Find $\displaystyle{\lim_{x \to \, -\infty}\left(x + \sqrt{x^2 + ax} \right) }$, where $a$ is a constant.

Click to View Calculus Solution

As $x \to \, -\infty$, the $x$-term becomes larger and larger in the negative direction, while the square-root term becomes larger and larger in the positive direciton. We thus don’t immediately know what the difference between the two terms is. $(“-\infty + \infty”$ 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 $x – \sqrt{x^2 + ax}$ divided by itself:

\[ \begin{align*}
\lim_{x \to \, -\infty}\left(x + \sqrt{x^2 + ax} \right) &= \lim_{x \to \, -\infty}\left(\frac{x + \sqrt{x^2 + ax}}{1} \right) \cdot \frac{x – \sqrt{x^2 + ax}}{x – \sqrt{x^2 + ax}} \\[8px] &= \lim_{x \to\, -\infty} \frac{x^2 – x \sqrt{x^2 + ax} + x \sqrt{x^2 + ax} – \left(\sqrt{x^2 + ax} \right)^2}{x – \sqrt{x^2 + ax}}\\[8px] &= \lim_{x \to \,-\infty} \frac{x^2 -\left(x^2 + ax \right)^2}{x – \sqrt{x^2 + ax}} \\[8px] &= \lim_{x \to\, -\infty} \frac{-ax}{x – \sqrt{x^2 + ax}}
\end{align*} \]

Since we’re looking at $x \to\, -\infty$ we’re interested only in negative values of $x$, and so we have $x = -\sqrt{x^2}$.
\[ \begin{align*}
\phantom{x + \sqrt{x^2 + ax} } &= \lim_{x \to \, -\infty} \frac{\dfrac{-ax}{x}}{\dfrac{x – \sqrt{x^2 + ax}}{x}}\\[8px] &= \lim_{x \to \, -\infty} \frac{-a}{\dfrac{x}{x} – \dfrac{\sqrt{x^2 + ax}}{-\sqrt{x^2}}}\\[8px] &= \lim_{x \to\, -\infty}\frac{-a}{1 + \sqrt{1 + \frac{a}{x}}} \\[8px] &= \frac{-a}{1 + \sqrt{1}} \\[8px] &= \frac{-a}{2} \quad \cmark
\end{align*} \]

Note that in the second to last line, we used the fact that $\displaystyle{\lim_{x \to\, -\infty} \frac{a}{x} = 0 }.$

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