## Limit at Infinity Problems with Square Roots

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nev
8 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

Matheno
Editor
8 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.

Anonymous
1 year ago

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

Matheno
Editor
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 Jayshree 2 years ago Please help Matheno Editor Reply to Jayshree 2 years 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 2 years ago √x+√x+√x/√x please help Last edited 2 years ago by Jayshree Matheno Editor Reply to Jayshree 2 years 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 2 years ago root of x ( root of x+c – root x ) at lim x tends to infinite 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]