Matheno - Learn Well and Excel

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.

[collapse]
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

[collapse]
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:
\[ \begin{align*}
f(x) &= \sqrt{x + \sqrt{x}} – \sqrt{x} \\[8px] f(100) &= \sqrt{100 + \sqrt{100}} – \sqrt{100} \approx 10.48 – 10 = 0.48 \\[8px] f(10,000) &= \sqrt{10,000 + 100} – 100 \approx 100.499 – 100 = 0.499 \\[8px] \end{align*} \]

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

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

[collapse]
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 $\displaystyle{\lim_{x \to\, -\infty} \frac{a}{x} = 0 }.$

[collapse]
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*}

[collapse]
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! : )

[collapse]
Small owl logo
Want access to all of our Calculus problems and solutions? Buy full access now — it’s quick and easy!



Subscribe
I'd like to be
8 Comments
newest
oldest most voted
Inline Feedbacks
View all comments
Anonymous
1 month ago

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

Matheno
Editor
Reply to  Anonymous
1 month 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 month ago by Matheno
Jayshree
4 months ago

Please help

Matheno
Editor
Reply to  Jayshree
4 months 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!

Jayshree
4 months ago

√x+√x+√x/√x please help

Last edited 4 months ago by Jayshree
Matheno
Editor
Reply to  Jayshree
4 months 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!

Anonymous
10 months ago

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

Matheno
Editor
Reply to  Anonymous
10 months 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.

Anonymous
1 year ago

√x√x√x-√x Lim app to infinity is

Matheno
Editor
Reply to  Anonymous
1 year ago

We’ve updated the page to provide the solution to this question as Practice Problem #7 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!