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At the institution where I teach, regrettably, we do not teach students the real definition of the limit in our calculus classes. I am teaching a little complex analysis, and I would like to use the notion of $\lim_{z \to z_0} f(z)$, with at least a little rigor. If I mention $\epsilon$ and $\delta$ my students will not understand it at all. Is $\lim_{z \to z_0} f(z) =v$ equivalent to the assertion that $f(r(t)) \to v$ as $t:0\to 1$ along any path $r:[0,1) \to \mathbb{C}$ with $r(t) \to z_0$ as $t \to 1^-$? My students might appreciate the idea of the pathwise definition. It is clear to me that the standard, $\epsilon$-$\delta$ definition implies the pathwise criterion, but I embarassed to admit that I do not know if the converse is true.

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Self-promotion: How much multivariable calculus can be done along curves?. The proof for continuity given there works for limits as well. – user31373 Sep 27 '12 at 3:21
Thank you. I am embarassed that the proof is so easy. If you give your response as an answer I'll upvote it. – Stefan Smith Sep 27 '12 at 12:48
up vote 4 down vote accepted

Proof by contrapositive. Suppose that $\lim_{z\to z_0}f(z)=v$ is false. Then there is a sequence $z_n\to z_0$ such that $f(z_n)\not\to v$. Define a path $r:[0,1]\to\mathbb C$ as follows: $r(1)=z_0$, $r(1-1/n)=z_{n}$ for each $n$, and $r$ is affine on each interval $[1-1/n,1-1/(n+1)]$. This is a continuous path ending at $z_0$, and $f(r(t))\not\to v$ as $t\to 1-$.

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