MHB Limit as x approaches negative infinity.

AI Thread Summary
The limit as x approaches negative infinity for the function is calculated as $$\lim_{{x}\to{-\infty}}\frac {x} {\sqrt{x^2}} = -1$$ because as x becomes increasingly negative, the expression simplifies to the sign function. The confusion arises from the square root of x squared, which is defined as the absolute value, leading to $$\sqrt{x^2} = |x|$$. For negative values of x, this results in $$|x| = -x$$, confirming that the limit evaluates to -1. Thus, the limit correctly reflects the behavior of the function as x approaches negative infinity.
tmt1
Messages
230
Reaction score
0
For this function:

$$\lim_{{x}\to{-\infty}}\frac {x} {\sqrt{x^2}} = -1$$

Why is this correct?

If x is equal to -1, for example, -1 square is 1. And the square root of 1 is 1. So should the answer be 1?
 
Mathematics news on Phys.org
If $x=-1$, then you have:

$$\frac{-1}{1}=-1$$

The expression in the limit is one way to define the sign function:

$$\text{sgn}(x)\equiv\frac{x}{|x|}$$ where $$x\ne0$$

This is equivalent to the piecewise definition:

$$\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}$$
 
MarkFL said:
If $x=-1$, then you have:

$$\frac{-1}{1}=-1$$

The expression in the limit is one way to define the sign function:

$$\text{sgn}(x)\equiv\frac{x}{|x|}$$ where $$x\ne0$$

This is equivalent to the piecewise definition:

$$\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}$$

so $$\sqrt{ (-x)^2} = x$$?
 
tmt said:
so $$\sqrt{ (-x)^2} = x$$?

No, what we have is:

$$\sqrt{(-x)^2}=\sqrt{x^2}\equiv|x|$$

Recall that:

$$|x|\equiv\begin{cases}-x, & x<0 \\[3pt] x, & 0\le x \\ \end{cases}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top