Limits with Square root

Question

I have this ridiculous doubt, why don't we take $+$ or $-$ while taking the square root of a number at the last step of finding a limit?

For example, $$\lim_{x\to 0} \frac{\sqrt{1-x^2}- \sqrt{1+x^2}}{x^2}$$

After rationalising and applying the limit, we get: $$\frac{-2}{\sqrt1+\sqrt1}$$

In this last step, why don't we take $4$ different cases for the two $\sqrt1$, like $(+,+), (+,-), (-,+)$ and $(-,-)$? Rather everyone takes only the positive case for the square roots.

Not just in this case, in all limits with square roots, we take only the positive case of square root.

Welcome to MSE. $\sqrt x$ usually denotes the non-negative square root — J. W. Tanner, Nov 17 '23 at 05:37
The square root function is always nonnegatively valued. The $\pm$ enters when solving an equation with squares in it. I.e., $x^2=4$ means $x=\pm 2$, but $\sqrt 4$ is just $2$. — pancini, Nov 17 '23 at 05:38
Thanks all for the replies. I didn't think of square root as a function. — Adarsh Dev, Nov 17 '23 at 05:54
$4$ has two square roots, the one denoted by an "r", that is $\sqrt{\cdot}$ is the non-negative one. $\sqrt{4}=2$. — GReyes, Nov 17 '23 at 07:00
https://math.stackexchange.com/q/26363/925076 ,Might be helpful — Dheeraj Gujrathi, Nov 17 '23 at 08:31

score 1 · Answer 1 · answered Nov 17 '23 at 08:26

In real analysis, unlike algebra, the square root is defined as a function $\sqrt{ }: [0, \infty) \rightarrow \mathbb{R}$ which outputs the POSITIVE square root of the given number (which is proven to be unique).

If we consider the negative square root among the outputs, the square root is no longer a function (since more than one output correspond to the same input), so the limit you've written doesn't make sense.

score 1 · Answer 2 · edited Nov 17 '23 at 10:36

By definition, $\sqrt{y^2}=|y|$

In case of the given limit, $$\lim_{x\to 0} \frac{\sqrt{1-x^2}- \sqrt{1+x^2}}{x^2}\\ {=\lim_{x\to 0} \frac{-2}{\sqrt{1-x^2}+ \sqrt{1+x^2}}\\ =\frac{-2}{\sqrt1+\sqrt1}}$$

Now, $$\sqrt{(\pm1)^2}=|\pm1|=1\ne-1\\$$

There is one only possible value of $\sqrt1$ not two. If we want a negetive number then you have to put '$-$' sign first. e.g. $-\sqrt1<0$

To learn more about these click here.

score 1 · Answer 3 · answered Nov 17 '23 at 10:30

By definition $\sqrt a$ is the nonnegative solution to the equation $x^2=a$. In other words, $\sqrt1=1$, even though the equation $x^2=1$ has two solutions.

In fact, by definition, the two solutions to $x^2=a$ are $x=\sqrt a$ and $x=-\sqrt a$. In other words, the $\pm$ comes in front of the $\sqrt{}$ symbol precisely because the number $\sqrt{a}$ is always positive.

Limits with Square root

3 Answers3