Limits for $\sqrt{x^2}-x$ as $x \to \infty$ or $x \to -\infty$ I know what the final answer should be since $\sqrt{x^2}=|x|$ and so the function becomes $|x|-x=0$ for positive numbers and $-x-x=-2x$ for negative numbers.
But I'm just trying to do the limit with multiplying and dividing by the conjugate $\frac{\sqrt{x^2}+x}{\sqrt{x^2}+x}$ and I can't seem to make it work out.
I would get something that looks like $\frac{(\sqrt{x^2})^2 - (x)^2}{\sqrt{x^2} + x} $ and then I should simplify something as $x$ based on $x>0$ or $-x$ based on $x<0$
For the positive case, I get $\frac{x^2-x^2}{x+x} = 0$
For the negative case, I get $\frac{(-x)^2-x^2}{-x+x}$ and I not sure if I missed something painfully obvious but I'm not sure how I could simplify this to get $-2x$

*

*The reason I focused on conjugates is because that seems to be how most 1st year questions of this format seem to operate but I wanted to make my own questions and see if conjugates would always work

 A: This question cannot be done by "conjugation".
To see why, one has to recall what "conjugation" is. At the level of what the author here seems to know (one can actually do better, but whatever), conjugation is the usage of the identities $$
a-b = \frac{a^2-b^2}{a+b} \quad ; \quad a+b = \frac{a^2-b^2}{a-b}
$$
to either simplify limit expressions containing radicals (typically : sometimes it's used for trigonometric expressions as well) or eliminate singularities such as $\frac{0}{0}$ expressions.
For example, take $\lim_{n \to \infty} (\sqrt{n+1} - \sqrt{n})$. One cannot write this as $\lim_{n \to \infty} \sqrt{n+1} - \lim_{n \to \infty} \sqrt{n}$ because limit laws don't allow it. Conjugation enters the picture : $$
\sqrt{n+1} - \sqrt{n} = \frac{(n+1)-n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}
$$
It becomes fairly clear what the limit is by analyzing the denominator. The limit value may not have been as evident from the original expression.

Conjugation is fairly limited as a technique. For starters, you need an expression of the form $a-b$ to transform, and if there's more than three terms that you want to simplify together then you're done for.
Next, there's no reason why, in general, $\frac{a^2-b^2}{a+b}$ is any simpler than $a-b$, or that $\frac{a^2-b^2}{a-b}$ is any simpler than $a+b$. For example, if $a = \sin^2 \theta$ and $b = \cos^2 \theta$, then $$
a+b =\sin^2 \theta + \cos^2 \theta =  1 \quad ; \quad \frac{a^2-b^2}{a-b} = \frac{\sin^4 \theta - \cos^4 \theta}{ \sin^2 \theta - \cos^2 \theta}
$$
It is obvious which one is a simpler expression. Typically, conjugation is performed when it is observed that $a^2-b^2$ (which is expected to be the most "complicated term") is an extremely simple quantity.
However, there's a third realm in which conjugation falls apart.

The identity $$
x+y = \frac{x^2-y^2}{x-y}
$$
is not well defined if $x=y$. Similarly,$$
x-y = \frac{x^2-y^2}{x+y}
$$
is not well-defined if $x+y=0$.

That is , one cannot use conjugation when it turns out that the denominator of the right hand side of the conjugation is zero. That's what is happening in our example.
--
Take , say $$
\lim_{x \to 0} (x+x)
$$
For starters, $x+x = 2x$, so this limit is just $0$. However, anybody writing $$
x+x = \frac{(x+x)(x-x)}{(x-x)} = \frac{x^2-x^2}{x-x}
$$
and asking "$\frac{0}{0}$, where have I gone wrong" is definitely going to have to be told to stare at the LHS more carefully. This manipulation is literally wrong for every value of $x$, because $x-x=0$ for every value of $x$!
Conjugation will absolutely fall apart when you end up dividing by an expression that is identically zero near the point where the limit is being taken.

The same is happening here. Indeed, you have $$
\sqrt{x^2}-x
$$
for $x<0$. In that case, if we were to try conjugation, we'd land up with $$
\sqrt{x^2}-x = \frac{(\sqrt{x^2})^2 - x^2}{\sqrt{x^2}+x} 
$$
This manipulation is wrong, because $\sqrt{x^2}+x =0$ for $x<0$. Therefore, we see that conjugation fails.
However, conjugation doesn't fail from a simple point of view. Say you tried to manipulate the expressions so that you can perform conjugation. The point is : you can't. Say you try adding and subtracting $a$. Then you get $(\sqrt{x^2}+a) - (x+a)$ , but these two bracketed terms are still equal so conjugation can't happen. Ditto if you tried to multiply and divide $x$ by some $a$, for example.
Therefore, this one particular case is pretty much immune to conjugation. However, it's obviously solvable because $\sqrt{x^2} = -x$ for $x<0$, so that $\sqrt{x^2}-x = -2x$, which leads to the answer $+\infty$.

To summarize : you don't want to use conjugation on a term like $a-b$ if :

*

*$a-b$ is part of a larger group of terms $a-b+c+d\ldots$ which you want to simplify in one stroke. This isn't wrong, but it will lead you down a complicated path. You can still do this by grouping as I show later.


*$a^2-b^2$ is a more complicated term than $a+b$ or $a-b$. This will just complicate things.


*$a=b$ or $a=-b$. Conjugation is plain mathematically wrong in this scenario (although, one could proceed down it if one eventually wishes to contradict limit existence, for example)

I've absolutely destroyed conjugation as a method in the previous sections. It's reputation is lying in tatters. Therefore, we should talk about when it is useful : that is true in a LOT of cases as well. So when can you use conjugation?

*

*The sum of radicals, or of a radical and a non-radical, which can be either in a numerator or in a denominator (or in both!). See here. This calculation is actually the limit of a differential quotient : so it's important because it will appear in the first-principle calculation of a derivative. See here where conjugation is used to simplify both a numerator and a denominator.


*In a $\frac{0}{0}$ calculation, where you get the feeling that terms can cancel out. Over here, the example used is similar to the previous ones in that all involve a $\frac{0}{0}$, but different in that the $x-4$ is actually a "multiple" of $\sqrt{x}-2$, so conjugation is done with the idea that the $a^2-b^2 = x-4$ will cancel with the denominator, rather than be something simple like , say, $1$.


*Sometimes, trigonometric identities such as $\sec^2 x - \tan^2 x = 1$ show that terms of the form $a^2-b^2$ can be simple when $a,b$ are trigonometric quantities. Therefore, here you can find one example of conjugation in the trigonometric setting, along with a generalized version of conjugation known as rationalization (which can be used for multiple surds, or for higher order surds).

Finally, there needs to be no real insistence on $x$ going to "minus infinity". Simply writing $y = -x$ and reframing the limit in terms of $-y$ provides a limit that will go to $+\infty$ instead.
Let's do that for the given case. Given $\sqrt{x^2} - x$ as $x \to -\infty$, we take $y = -x$ so that $y \to +\infty$ now.
This gives us $\sqrt{(-y)^2} - (-y) = \sqrt{y^2}+y$. Can we apply conjugation to this? NO, because if $\sqrt{y^2}=a$ and $y=b$ then $a=b$, and in that case, $a-b=0$ so you can't divide by $a-b$.
Therefore, there's no real need to insist on studying $-\infty$ as a separate variable limit : if conjugation doesn't work, then it doesn't do so independent of ordinary limit reformulations which change the variable limit.
A: $$f(x)=\sqrt{x^2}-x=|x|-x$$
$$\lim_{x\to \infty} f(x)=\lim_{x \to \infty} \frac{|x|^2-x^2}{|x|+x}=\frac{0}{\infty}=0$$
$$\lim_{x\to -\infty} f(x)=\lim_{x \to -\infty} -x-x= \lim_{x\to -\infty} -2x=\infty$$
A: As noticed, "conjugation" in this case is fine when $x>0$ but it doesn't work when $x<0$ since the quantity $\sqrt{x^2}+x=0$ which leads to an undefined expression $\frac 0 0$.
The latter is the key point and nothing else needs to be added.

To use "conjugation" at all costs, in order to eliminate the aforementioned issue, we could consider the "equivalent" limit obtained adding the vanishing term $\frac1x \to 0$
$$\lim_{x\to -\infty}\sqrt{x^2}-x+\frac1x$$
such that the "conjugate quantity" $\sqrt{x^2}+x-\frac1x\neq 0$ to obtain
$$\sqrt{x^2}-x+\frac1x=\left(\sqrt{x^2}-x+\frac1x\right) \frac{\sqrt{x^2}+x-\frac1x}{\sqrt{x^2}+x-\frac1x}=\frac{\left(\sqrt{x^2}\right)^2-\left(x-\frac1x\right)^2}{\sqrt{x^2}+x-\frac1x}=$$
$$=\frac{x^2-x^2-\frac1{x^2}+2}{\sqrt{x^2}+x-\frac1x}=\frac{2-\frac1{x^2}}{-\frac1x}=-2x+\frac1x \to \infty$$
which maybe is not an effective approach but at least is valid and allows to conclude that
$$\lim_{x\to -\infty}\sqrt{x^2}-x=\infty$$
