I'm trying to prove this lemma, but I'm stumped. This lemma is presented first thing after the definition of a limit for complex functions, so I don't have previous results to use here.  The left hand statement translates to:  $\forall \varepsilon &gt; 0, \exists \delta &gt; 0$ such that $0 &lt; |z - a| &lt; \delta \implies |f(z) - L| &lt; \varepsilon$  Right hand statement translates to:  $\forall (z_n)$ s.t. $z_n \to a$, $f(z_n) \to L$  Any guidance would be appreciated
I want to prove the definition of limit of a function from definition (1) to definition (2).  (1) A function $f(x)$ is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for every sequence $\{x_n\}$ with limit $x^*$, the limit $\lim_{n-&gt;\infty}f(x_n)$ exists and is equal to $f^*$. (2) A function is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for each $\epsilon &gt;0$, there exists a $\delta_\epsilon &gt;0$ such that $|f(x)-f^*| &lt;\epsilon \forall |x^*-x|&lt;\delta_\epsilon$  I am confused about the steps. Suppose I am given an $\epsilon &gt; 0$ and $x^*$. I have to show the existence of a $\delta_\epsilon &gt;0$ such that for all $x$ satisfying  $|x^*-x|&lt;\delta_\epsilon$ I can construct a sequence $x_n$ which converges to $x^*$ and $x_n=x$ for some $n \geq N$. Then I can say that $f(x_n)-f^*=f(x)-f^* &lt; \epsilon$. I think I may have made some mistake about the steps.  My main confusion is the following : Suppose I choose $\delta_\epsilon=\epsilon$. Then if a sequence $\{x_n\}$ converges to $x^*$ then $x_n=x$ (where $x$ satisfies $|x-x^*|&lt;\epsilon$) for some $n\geq N$. Now, $\lim_{n-&gt;\infty}f(x_n)$ exists implies there exists an $M$ such that $\forall n\geq M$, $f(x_n)-f^* &lt;\epsilon$. Now, I need to show that for some $n \geq M$ indeed $x_n=x$. How do I show that ? This $M$ and $N$ are different.
LaTeX seems to be having problems handling nested optional arguments. I've included a minimal code example that generates the error:  \documentclass[12pt,oneside]{book} \begin{document}  \newcommand{\termbad}[2][]{#2#1} \newcommand{\termgood}[2][]{#2#1}  \newcommand{\up}[1]{#1} \newcommand{\dd}[1][]{#1}  \termgood{\up{Tart}}{Pop} \termbad[\up{Tart}]{Pop}      \termgood{\dd[Tart]}{Pop} \termbad[{\dd[Tart]}]{Pop} %\termbad[\dd[Tart]]{Pop}   \end{document}   If you run this code, it works. If you remove the % in the second to last line, it fails with:  Runaway argument?  Tart{Pop} \par \par \end {document}  !File ended while scanning use of \dd.   After further tinkering, I have discovered the problem: LaTeX is reading the ] for the inner optional argument as the end of the outer optional argument and then passing a malformed argument in. So it is passing in \dd[Tart as the optional argument to \termbad instead of \dd[Tart].  This still feels like a bug to me. 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