Why is x more than and equal to 0 and not -x less than and equal to 0 instead? When (x approach 0- ) for lim | x | divide by x .... hope someone could answer this T.T

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txmedteach's profile pic

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

It sounds like you're asking about the following limit:

`lim_(x->0+) |x|/x`

You have to know whether you're approaching from the right or left based on what the problem asks because the limit is different depending on which you choose.

From the rightx is positive, so |x| = x, giving us the following simplification:

`lim_(x->0+) |x|/x = lim_(x->0+) x/x = 1`

If you take the limit from the leftx is negative, so |x| = -x, giving us a different result:

`lim_(x->0-) |x|/x = lim_(x->0-) (-x)/x = -1`

As you can see, choosing whether to approach zero from the right or left is an important distinction with this limit. If you were to simply ask the solution to the the limit from both sides, the answer would be simple from the above: the two-sided limit does not exist. Recall, in order for the unconstrained limit to exist, the limit from the left must be equal to the limit from the right.

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The limit `lim_(x->0-) (|x|)/x` has to be determined.

`x->0-` indicates that the value of x approaches 0 from the side where the values of x are less than 0.

If `x->0-` , `-x->0+` ; the two are equivalent.

When `x-> 0-` , the value of x is negative, , the limit `lim_(x->0-)(|x|)/x` = `lim_(x->0-)x/(-x)` = `lim_(x->0-) -1 = -1`

Calculating the same by using `-x->0+` , gives the limit `lim_(-x->0+)(|-x|)/(-x)`

For values of x greater than 0, (-x) is negative.

`lim_(-x->0+)(|-x|)/(-x)`

= `lim_(-x->0+)(x)/(-x)`

= `lim_(-x->0+) -1`

= -1

The limit `lim_(x->0-) (|x|)/x = -1`

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