Precise Definition of a Limit at Negative Infinity

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SUMMARY

The discussion focuses on the formal definition of limits as \( x \) approaches negative infinity, specifically \( \lim_{x \to -\infty} f(x) = L \). The proposed definition states that this limit holds if for every \( \epsilon > 0 \), there exists an \( N \) such that if \( x < N \), then \( |f(x) - L| < \epsilon \). This definition is confirmed as correct by participants in the forum, clarifying that the condition \( x < N \) is appropriate for limits approaching negative infinity.

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  • Understanding of limits in calculus
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  • Knowledge of Stewart's Calculus, 6th Edition
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Tsunoyukami
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I'm working through some problems from Stewart's Calulus, 6ed. and am having some difficulty with certain limit proofs. In particular, there is no definition provided for limits of the form:

$$ \lim_{x \to - \infty} f(x) = L $$

One of the exercises is to come up with a formal definition for such a proof - which I did and used successfully a few times about a month ago when I was working through that section - but I want to make sure that my definition is really correct and not just lucky for those two problems.


I would suspect the definition of such a limit to be something like:

We say

$$ \lim_{x \to - \infty} f(x) = L $$

if for every ##\epsilon>0## there exists ##N## such that ##x < N \implies |f(x) - L| < \epsilon##.


Is this the correct precise definition of such a limit? I was unable to find an answer browsing google - most websites seem to provide the precise definition for "normal" limits and for limits at positive infinity but not at negative infinity.
 
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Sounds good.
 
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Thank you for your prompt reply!
 
Would sound even better if you had written ##x > N##, but I'm pretty sure the < was just a typo...
 
BvU said:
Would sound even better if you had written ##x > N##, but I'm pretty sure the < was just a typo...
The OP wrote it correctly because the limit is for ##x## going to ##-\infty##.
 
My apologies. Couldn't imagine such a quirk; proves how rusty one can get with old age... :redface:
 

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