Are these limits equivalent? $$ \lim\limits_{x \rightarrow p}f(x) \\\lim\limits_{h \rightarrow 0}f(p+h) $$ They both describe $f(x)$ as $x$ gets arbitrarily close to $p$, but I see both used in different contexts so I'm wondering if there is actually a difference between the two. Thanks!
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1Do you mean to have the second limit be $\lim_{h \to 0^+}{f(p+h)}$? – Hayden Mar 19 '21 at 06:41
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@Hayden. I changed the question, hopefully it makes sense now. This problem from baby Rudin is what lead me ask this question. – Henry Mar 19 '21 at 06:45
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1no deep maths facts here. Just make the algebraic substitution x=p+h – Benjamin Wang Mar 19 '21 at 06:53
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1They're exactly the same. It's just that sometimes one of the two is easier to do calculations with. – Vercassivelaunos Mar 19 '21 at 06:53
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They are actually the same.
Let $\lim\limits_{x \rightarrow p}f(x)=l\in \mathbb R$. What existence of this limit means is that for every $\epsilon\gt 0,\exists \delta\gt 0: 0\lt |x-p|\lt \delta\implies |f(x)-l|\lt \epsilon$
WLOG, put $x=p+h$ so that we have $0\lt |h| \lt \delta\implies |f(p+h)-l|\lt \epsilon$
Now use definition of limit again to finish the proof.
Koro
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