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
\[ \bbox[yellow,5px]{
\begin{align*}
\text{If $x$ is positive: } x &= \sqrt{x^2} \\[8px]
\text{If $x$ is negative: } x &= -\sqrt{x^2} \\[8px]
\end{align*} } \]
• By contrast, if $x = -3$, then $x = -3 = -\sqrt{9}$.
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.
(b) Find $\displaystyle{\lim_{x \to\, -\infty}\frac{\sqrt{5x^2 + 2x}}{x}}.$
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 $\displaystyle{\lim_{x \to \infty}\frac{2}{x} = 0}$.
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.
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.
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.
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{x^2 + x} – x \right)}.$
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:
\[ \begin{align*}
f(x) &= \sqrt{x^2 + x} – x \\[8px]
f(10) &= \sqrt{100 + 10} – 10 \approx 10.488 – 10 = 0.488 \\[8px]
f(20) &= \sqrt{400 + 20} – 20 \approx 20.494 – 20 = 0.494 \\[8px]
f(100) &= \sqrt{10,000 + 100} – 100 \approx 100.499 -100 = 0.499
\end{align*} \]
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{x + \sqrt{x}} – \sqrt{x} \right)}$.
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:
\[ \begin{align*}
f(x) &= \sqrt{x + \sqrt{x}} – \sqrt{x} \\[8px]
f(100) &= \sqrt{100 + \sqrt{100}} – \sqrt{100} \approx 10.48 – 10 = 0.48 \\[8px]
f(10,000) &= \sqrt{10,000 + 100} – 100 \approx 100.499 – 100 = 0.499 \\[8px]
\end{align*} \]
Find $\displaystyle{\lim_{x \to \infty}\left(\sqrt{a^2 x^2 + x} -ax \right)},$ where $a$ is a positive constant.
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 $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$
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 $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}.$
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 $\displaystyle{\lim_{x \to\, -\infty} \frac{a}{x} = 0 }.$
Find $\displaystyle{\lim_{x \to \infty}\big[\sqrt{x}\sqrt{x}\sqrt{x} – \sqrt{x} \big]}.$
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*}
Find $\displaystyle{ \lim_{x \to \infty}\sqrt{x + \sqrt{x + \sqrt{x}}} – \sqrt{x} }$.
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! : )

Want access to all of our Calculus problems and solutions? Buy full access now — it’s quick and easy!
limit of square root of x^2+1 – square root of x+1 as x approaches infinity
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!
Please help
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!
√x+√x+√x/√x please help
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!
root of x ( root of x+c – root x ) at lim x tends to infinite
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.
√x√x√x-√x Lim app to infinity is
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!