Why is the limit at abs(x)/x as x -> 0 not exist. As it approaches infinity does the limit equal 1?

Question

limit |x|
x->0 __
x Why does it not exist? what about as x-> infinity?

limit |x|
x->0 __
. x

limit |x|
x->0 __
..........x

Answer 1

What you have there is basically a version of the "discontinuous sign function," sgn(x). This function takes any real nonzero number, and returns either +1, or -1 depending on whether the argument is positive or negative.

Since you can't divide by zero, |x| / x is undefined when x = 0. In an abstract sense, sgn (x) is undefined at zero, because zero is neither negative nor positive.

Furthermore, this function does not have a limit, because as you approach 0 from the negative side, the function is -1, but as you approach from the positive side, it's value is +1.

You can confirm this using L'Hospital's Rule:

lim x-->a {u(x) / v(x)} = { lim x-->a (du/dx) / lim x-->a (dv/dx)}

The derivative of x is just 1, so no problem with dividing by zero there. But the derivative of |x| IS a problem. In fact,

d |x| / dx = sgn(x);

therefore, the assumption that the limit of sgn(x) exists at zero leads to a paradox. The only acceptable possibility is that the limit does not exist.Q.e.D.

~W.O.M.B.A.T.

Answer 2

The limit of abs(x)/x as x-->0 does not exist because the left- and right-hand limits are not equal. From the left, the limit is -1; from the right it is +1.

As x-->infinity, then the limit is +1...but the limit is still -1 if x--> -infinity.

Answer 3

The limit as it approaches zero is |0| over zero, and anything over zero isn't defined until later, if then. Here's why
1. Does 0/0 = 1? It is divided by itself, and x/x = 1, right?
2. Does it equal 0? 0/X = 0, like zero over anything, right?
Zero is a special case.

Later in calculus you'll find there is a difference between +0, and -0, and more considerations, but I don't think you're there yet?

As x approaches anything positive the limit approaches 1, and anything negative, -1. So going from (-inf to -0), the limit is -1, then at 0 it is discontinuous, then at (+0 to inf) it is 1. See that weird switchover at zero - that's the issue.

Answer 4

TA is right ALWAYS ;) if you plug in x=0 you get [0/0] which is an indeterminate. Apply L'Hop rule: lim x => 0 (2x) / -sinx ===>still [0/0] repeat L'Hop lim x=> 0 (2) / -cosx ===> finally 2/(-1) = -1 ANSWER = -2 Also, you should know lim x=>0 of sinx / x = x /sinx both ==> zero if you do not know L'Hopital's rule, then multiply the numerator and denominator by the conjugate: (x^2) / (cosx -1) multiply top/bottom by (cosx +1) x^2 (cosx + 1) / (cos^2(x) - 1) ====> use Pyth Id sin^2x+cos^2x=1 =x^2(cosx + 1) / -sin^2(x) = - ( x/sinx )( x/sinx ) (cosx + 1) Each sinx / x limit goes to one as x goes to zero hence = - (cos x + 1) = -(cos0 + 1) = -(1+1) = -2 :)

Answer 5

The direction of the limit is not defined as positive or negative.

So there isn't one.

Answer 6

because infinity over infinity doesnt equal 1. it can equal almost anything you want it to equal.