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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 iconYou 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.

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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.)

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 $\sqrt{x^2 + x} + x$ divided by itself:

\[ \begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + x} – x \right) &= \lim_{x \to \infty}\left(\frac{\sqrt{x^2 + x} – x}{1} \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}\right) \\[8px] &= \lim_{x \to \infty}\frac{\left( \sqrt{x^2 + x}\right)^2 + x\sqrt{x^2 + x} -x \sqrt{x^2 + x} -x^2 }{\sqrt{x^2 + x} + x} \\[8px] &= \lim_{x \to \infty}\frac{(x^2 + x) -x^2}{\sqrt{x^2 + x} + x} \\[8px] &= \lim_{x \to \infty}\frac{x}{\sqrt{x^2 + x} + x}
\end{align*} \] 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:
\[ \begin{align*}
f(x) &= \sqrt{x^2 + x} – x \\[8px] f(10) &= \sqrt{100 + 10} – 10 \approx 10.488 – 10 = 0.488 \\[8px] f(20) &= \sqrt{400 + 20} – 20 \approx 20.494 – 20 = 0.494 \\[8px] f(100) &= \sqrt{10,000 + 100} – 100 \approx 100.499 -100 = 0.499
\end{align*} \]

As x goes to infinity, the curve y = f(x) approaches y = 1/2

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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.)

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 $\sqrt{x + \sqrt{x}} + \sqrt{x}$ divided by itself:

\[ \begin{align*}
\lim_{x \to \infty}\left(\sqrt{x + \sqrt{x}} – \sqrt{x} \right) &= \lim_{x \to \infty}\left(\frac{\sqrt{x + \sqrt{x}} – \sqrt{x}}{1} \cdot \frac{\sqrt{x + \sqrt{x}} + \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \right) \\[8px] &= \lim_{x \to \infty}\frac{\left(\sqrt{x + \sqrt{x}} \right)^2 – \sqrt{x}\sqrt{x + \sqrt{x}} + \sqrt{x}\sqrt{x + \sqrt{x}} – \left( \sqrt{x}\right)^2}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty} \frac{\left(x + \sqrt{x} \right) – x}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x}} \\[8px] \end{align*} \] 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:
\[ \begin{align*}
f(x) &= \sqrt{x + \sqrt{x}} – \sqrt{x} \\[8px] f(100) &= \sqrt{100 + \sqrt{100}} – \sqrt{100} \approx 10.48 – 10 = 0.48 \\[8px] f(10,000) &= \sqrt{10,000 + 100} – 100 \approx 100.499 – 100 = 0.499 \\[8px] \end{align*} \]

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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.)

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 $\sqrt{a^2 x^2 + x} + ax$ divided by itself:

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

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 $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$

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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 $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$

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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:

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

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 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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Practice Problem #7
A student posted this question in the Comments below.
Find $\displaystyle{\lim_{x \to \infty}\big[\sqrt{x}\sqrt{x}\sqrt{x} – \sqrt{x} \big]}.$
Click to View Calculus Solution
Since $ \displaystyle{\lim_{x \to \infty}\sqrt{x^3}} = \infty$ and $ \displaystyle{\lim_{x \to \infty}\sqrt{x}} = \infty,$ we have $”\infty – \infty”$ which is indeterminate: we don’t know what the limit actually is, only that we have more work to do.

As is so often the case, factoring provides a way forward: notice that we can pull a $\sqrt{x}$ out of both terms:

\begin{align*}
\sqrt{x}\sqrt{x}\sqrt{x} – \sqrt{x} &= \sqrt{x} \big( \sqrt{x}\sqrt{x} – 1 \big) \\[8px] &= \sqrt{x}(x – 1)
\end{align*}
Once we’ve done that, the limit becomes clear:
\begin{align*}
\lim_{x \to \infty}\big[\sqrt{x}\sqrt{x}\sqrt{x} – \sqrt{x} \big] &= \lim_{x \to \infty}\big[\sqrt{x}(x – 1)\big] \\[8px] &= \big[\lim_{x \to \infty}\sqrt{x} \big] \cdot \big[\lim_{x \to \infty}(x – 1)\big] \\[8px] &= \infty \cdot \infty \\[8px] &= \infty \quad \cmark
\end{align*}

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Practice Problem #8
A student posted this question in the Comments below.
Find $\displaystyle{ \lim_{x \to \infty}\sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x} }$.
Click to View Calculus Solution
There are two main steps to find this limit.

Step 1: As we have in the problems above, we multiply the expression by its conjugate divided by itself:
\begin{align*}
\lim_{x \to \infty}\left( \sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x}\right) &= \lim_{x \to \infty}\left(\sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x} \right)\cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty} \frac{\left(\sqrt{x + \sqrt{x + \sqrt{x}}}\right)\left(\sqrt{x + \sqrt{x + \sqrt{x}}} \right) + \cancel{\left(\sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x} \right)\left(\sqrt{x} \right)} – \cancel{\left( \sqrt{x}\right)\left(\sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x} \right)} + \left( -\sqrt{x}\right)\left(\sqrt{x} \right) }{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}\\[8px] &=\lim_{x \to \infty} \frac{\left(x + \sqrt{x + \sqrt{x}} \right) -x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}
\end{align*}
With this rewritten expression, you might be able to look at it and see that the numerator is dominated by the (first) $\sqrt{x}$ term, while the numerator is dominated equally by two factors of $\sqrt{x}$, and so the limit will be $\dfrac{1}{2}.$

But in case you don’t see this (because you haven’t yet done 10,000 of these types of problems), or you need to prove it, we move to our usual approach of Step 2: Divide the numerator and denominator both by the largest factor in the denominator, which is $\sqrt{x}.$ So let’s multiply both the numerator and denominator by $\dfrac{1}{\sqrt{x}}$:
\begin{align*}
\lim_{x \to \infty}\frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}
&= \lim_{x \to \infty}\frac{\frac{1}{\sqrt{x}}\sqrt{x + \sqrt{x}}}{\frac{1}{\sqrt{x}}\left( \sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}\right)} \\[8px] &= \lim_{x \to \infty}\frac{\sqrt{\frac{x + \sqrt{x}}{x}}}{ \sqrt{\frac{x + \sqrt{x + \sqrt{x}}}{x}} + \frac{\sqrt{x}}{\sqrt{x}}}\\[8px] &= \lim_{x \to \infty}\frac{\sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}}+1} \\[8px] &= \frac{\sqrt{1 + \cancelto{0}{ \left( \lim_{x \to \infty}\frac{1}{\sqrt{x}}\right)}}}{\sqrt{1 + \cancelto{0}{ \left( \lim_{x \to \infty}\frac{\sqrt{x + \sqrt{x}}}{x}\right)}}+1} \\[8px] &= \frac{\sqrt{1}}{\sqrt{1} + 1} \\[8px] &= \frac{1}{2} \quad \cmark
\end{align*}
Whew! : )

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nev
5 months ago

i dont understand how to choose between +/- values. since √4 for example is +/-2, you shouldnt be able to just throw a negative sign in front of the sqrt and treat the result of the sqrt to always be positive, i dont understand that– you would then have – +/-2. but it seems that if you dont treat sqrt’s this way, it messes up your evaluations… im lost

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Matheno
Editor
Reply to  nev
5 months ago

Hi, Nev,

This is a GREAT question. Thanks so much for asking! The short answer is: if you’re looking at a limit as x goes to negative infinity, you’ll always insert that negative sign “by hand.” Here’s why:

First, notice that when we’re looking at this limit as x goes to negative infinity, x takes on increasingly <em>negative</em> values. Crucially, x itself is always negative. 

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 naively 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 $x \to -\infty.$ 

This is a tricky issue — hence the reason we’re addressing it! We hope this additional discussion helps . . . and please let us know if it generates further questions for you, and we’ll do our best to address them.

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Anonymous
1 year ago

limit of square root of x^2+1 – square root of x+1 as x approaches infinity

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Matheno
Editor
Reply to  Anonymous
1 year ago

As with so many of the problems on this page, we first multiply by the conjugate of the given expression divided by itself:
\begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) \cdot \frac{\sqrt{x^2 + 1 } + \sqrt{x+1}}{\sqrt{x^2 + 1 } + \sqrt{x+1}} \\[8px]
&= \lim_{x \to \infty}\frac{\left(x^2 + 1\right) + \sqrt{x^2 + 1 }\sqrt{x+1} + \left(- \sqrt{x+1} \right)\sqrt{x^2 + 1 } – (x+1)}{\sqrt{x^2 + 1 } + \sqrt{x+1}} \\[8px]
&= \lim_{x \to \infty} \frac{x^2 -x}{\sqrt{x^2 + 1 } + \sqrt{x+1}}\\[8px]
\end{align*}
Next, we employ the “trick” we discussed on the page “Limits at Infinity,” and divide both the numerator and denominator by the largest power of x that is in the denominator. In this case, that power is $\sqrt{x^2} = x.$ Starting with the last line above:
\begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty} \frac{x^2 -x}{\sqrt{x^2 + 1 } + \sqrt{x+1}}\\[8px]
&= \lim_{x \to \infty} \frac{\dfrac{1}{x}\left(x^2 -x\right)}{\dfrac{1}{x}\left( \sqrt{x^2 + 1 } + \sqrt{x+1}\right)} \\[8px]
&= \lim_{x \to \infty} \frac{x-1}{\sqrt{\dfrac{x^2 + 1}{x^2} } + \sqrt{\dfrac{x+1}{x^2}}} \\[8px]
&= \lim_{x \to \infty} \frac{x-1}{\sqrt{1 + \dfrac{1}{x^2}} + \sqrt{\dfrac{1}{x} + \dfrac{1}{x^2}} }
\end{align*}
Now, $\displaystyle{\lim_{x \to \infty} \dfrac{1}{x^2} = 0 }$ and $\displaystyle{\lim_{x \to \infty} \dfrac{1}{x} = 0 }$, so
\begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2 + 1 } – \sqrt{x+1} \right) &= \lim_{x \to \infty} \frac{x-1}{\sqrt{1 + \cancelto{0}{ \dfrac{1}{x^2}}} + \sqrt{\cancelto{0}{ \dfrac{1}{x}} + \cancelto{0}{ \dfrac{1}{x^2}}} } \\[8px]
&= \lim_{x \to \infty} (x-1) = \infty \quad \cmark
\end{align*}
We hope that helps!

Last edited 1 year ago by Matheno
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Jayshree
1 year ago

Please help

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Matheno
Editor
Reply to  Jayshree
1 year ago

We’ve updated the page to provide the solution to this question as Practice Problem #8 above. It’s a good problem, and we hope we’ve been able to help you (and future students) solve it successfully for yourself.

Thanks for asking!

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Jayshree
1 year ago

√x+√x+√x/√x please help

Last edited 1 year ago by Jayshree
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Matheno
Editor
Reply to  Jayshree
1 year ago

We’ve updated the page to provide the solution to this question as Practice Problem #8 above. It’s a good problem, and we hope we’ve been able to help you (and future students) solve it successfully for yourself.

Thanks for asking!

0
Reply
Anonymous
2 years ago

root of x ( root of x+c – root x ) at lim x tends to infinite

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Matheno
Editor
Reply to  Anonymous
2 years ago

Thanks for asking! (Note that you may need to refresh your browser page for the math below to display properly.)

It turns out there are two strategic moves you have to make to find this limit.

First, as discussed above, any time we see a two square roots subtracted from each other, we automatically multiply by the conjugate
\[1= \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}}\]
So
\begin{align*}
\lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right) &= \lim_{x \to \infty} \sqrt{x}\left[\left( \sqrt{x+c} – \sqrt{x}\right) \cdot \left( \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}}\right) \right] \\[8px]
&= \lim_{x \to \infty}\sqrt{x} \left[ \frac{(x+c) +\sqrt{x+c}\sqrt{x} – \sqrt{x}\sqrt{x+c}-x }{\sqrt{x+c} + \sqrt{x}}\right] \\[8px]
&= \lim_{x \to \infty}\sqrt{x} \left[ \frac{c}{\sqrt{x+c} + \sqrt{x}}\right] \\[8px]
\end{align*}
The other strategic move we need to make is a “trick” we developed earlier : find the largest power of $x$ in the denominator, and then factor it out. Here, that power is $\sqrt{x}$:
\begin{align*}
\phantom{ \lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right)} &= \lim_{x \to \infty}\sqrt{x} \left[ \frac{c}{\sqrt{x}\left( \sqrt{1+\frac{c}{\sqrt{x}}} + 1\right)}\right] \\[8px]
&= \lim_{x \to \infty}\frac{\sqrt{x}}{\sqrt{x}} \left[ \frac{c}{\sqrt{1+\frac{c}{\sqrt{x}}} + 1}\right] \\[8px]
&= \lim_{x \to \infty} \frac{c}{ \sqrt{1+\frac{c}{\sqrt{x}}} + 1}
\end{align*}
Now, since $\lim_{x \to \infty} \dfrac{c}{\sqrt{x}} = 0,$ when we take the limit we have
\begin{align*}
\phantom{ \lim_{x \to \infty}\sqrt{x} \left( \sqrt{x+c} – \sqrt{x}\right)} &= \frac{c}{ \sqrt{1+\cancelto{0}{ \frac{c}{\sqrt{x}}}} + 1} \\[8px]
&= \frac{c}{2} \quad \cmark
\end{align*}
Whew! : )

We hope that helps.

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