"Why does x = − √x^2 when x < 0 due to the limit being as x approaches negative infinity?"

It isn't *due* to x approaching -∞.

x = −√x² when x < 0 *always* true.

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√x² has 2 values, +x and -x. But it is a (confusing) convention that “√A” is often taken to mean only the positive value of √A. If you want to indicate the two values you write them explicitly as +√A and -√A or ±√A. It follows that -√x² specifically means the negative root of x².

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x< 0, x = −√x² is always true. Just try it with any suitable (i.e. negative) value for x.

For example if x=-7

x² = 49.

√x² = √49 = ±7

We can write this as

+√x² = +7 (though + signs may be omitted)

-√x² = -7

We are only interested in the negative value.

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Why this is useful depends on the problem being solved but you haven't stated the problem.

I'm guessing you are finding a limit as x→-∞ and one of the terms in your equation is -√x². The hint is simply telling you that, in the limit, you can treat -√x² as equal to -∞.

Ohmygod

.......

x = −5 okay?

− √(x^2) <<<<<<<<<<<<<< claim is that this equals x for any negative x

= − √( (-5)^2 )

= − √25

= − 5

Which is x

Best wishes!

Many people take √(x²) to be equivalent to x. In fact, it is |x|. The principal square root of any real number must be non-negative.

Let x < 0.

√(x²) = |x|

√(x²) = -x ... because x < 0

-√(x²) = x

You can't divide anything by 0 and get a number or valid result.