# Limit x approach infinite Problem

Question: Prove that $\lim_{x\to\infty}f(x) = \lim_{x\to-\infty}f(-x)$

My Work:

If $\lim_{x\to\infty}f(x) = l$: by definition: $\forall \varepsilon > 0, \exists N$ such that if $x > N$, then $\vert f(x) - l \vert < \varepsilon$

Substitute $-x$ for $x$ that is $-x > N$, then $\vert f(-x) - l \vert < \varepsilon$

but $-x > N$ does not imply $x < N$ which is needed to satisfy the condition and show $\lim_{x\to-\infty}f(-x)$

What am I overlooking?

Thanks

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Hint: By the definition of limit as $x\to-\infty$, we want to show that for any $\epsilon \gt 0$, there exists an $M$ such that $|f(-x)-l|\lt \epsilon$ whenever $x\lt -M$.

Now let $M=N$, where the $N=N_\epsilon$ is as you described. Note that $x\gt N$ iff $-x\lt -N$.

An alternate definition of limit as $x\to-\infty$ is to ask that for every $\epsilon\gt 0$, there exists an $M$ such that $|g(u)-l|\lt \epsilon$ if $u\lt M$.

In that case, let $M=-N$. Note that $x\gt N$ iff $-x\lt -N$ iff $-x\lt M$.

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There seems to be a confusion of the definition. I thought for $\lim_{x\to-\infty}f(x) = l$ it was whenever $x < N$ and not $x < -N$? ... – mathnoob Oct 16 '12 at 3:17
It depends on who does the definition. If you want to use yours, let $M=-N$. – André Nicolas Oct 16 '12 at 3:21
Oh right because it is $\exists N$ so you can let it be whatever – mathnoob Oct 16 '12 at 3:22
@mathnoob: I have also written out the details for the definition you are using. Essentially no change. – André Nicolas Oct 16 '12 at 3:26
Ok thank you, I just confirmed the definition with someone else, they were using yours as well so maybe its preference. For this question, yours would be easier. – mathnoob Oct 16 '12 at 3:31