Limit question limit (x->0) 1/|x|

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Homework Help Overview

The discussion revolves around the limit of the function 1/|x| as x approaches 0, focusing on whether the limit exists and the implications of approaching from both the positive and negative sides.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the behavior of the limit from both sides of zero, questioning the definition of limits and the implications of approaching infinity.

Discussion Status

There is an ongoing examination of the limit's existence, with some participants asserting that the limit does not exist while others suggest that it can be interpreted as approaching +∞ under certain conditions. The conversation reflects differing interpretations of limit definitions.

Contextual Notes

Participants discuss the implications of using the extended real number line and the epsilon-delta definition in the context of limits approaching infinity.

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limit (x->0) 1/|x|
 
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the limit doesn't exist.
lim (x-> o+) = + ∞
lim (x-> o-) = - ∞

so limit of the function at the point zero is not existent by definition.
 
AdrianZ said:
the limit doesn't exist.
lim (x-> o+) = + ∞
lim (x-> o-) = - ∞

so limit of the function at the point zero is not existent by definition.

Not true. The function is 1/|x| not 1/x
 
Mentallic said:
Not true. The function is 1/|x| not 1/x

aah yea you're right. didn't notice the abs function there. anyway, when x approaches zero from left (and right) the function approaches +∞. so the limit is +∞.

that means for all positive deltas and M's if |x - 0| < delta, then f(x) > M. for any large M you like. intuitively you can say: the larger you want M to be, the smaller you need to take delta.
 
Last edited:
Which still means that there is no limit. Saying "[itex]\lim f(x)= \infty[/itex]" is just a way of saying that the limit does not exist for a particular reason.
 
well, I think if we're working on extended real number line then it doesn't hurt to say the limit of the function is +∞ because +∞ ∈ R*. and plus the epsilon, delta definition works fine with this example. correct me if I'm wrong.
 

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