Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem.  I have been losing lots of points for minor details that I've been unable to see.  Does the following proof hold?  Am I making any major (or minor) errors?
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right|+\frac{1}{3}<\epsilon\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon-\frac{1}{3}\\
&\implies \frac{1}{2} < \epsilon-\frac{1}{3}~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.)\\
&\implies 1<2(\epsilon-\frac{1}{3})\\
&\implies \left|x-1\right|<2\epsilon-\frac{2}{3}=\delta~~~~\mbox{(Because, choosing }x~s.t.~0<x<2\implies~-1<x-1<1)\\
\end{align*}
$\therefore \left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$

I think the following salvages the initial strategy while still holding true to the mathematical principals of epsilon-delta definition of limits.  Feedback welcome.  Please let me know if I have made another error.  I posted this as an answer so that the original question could remain for reference.
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon~~~~(|a|+|b|<\epsilon\implies|a|<\epsilon)\\
&\implies \frac{1}{2} < \epsilon~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.~\mbox{Hence, } \delta\leq 1)\\
&\implies 1<2\epsilon\\
&\implies |x-1|\leq\delta\implies-1\leq x-1\leq 1\implies 0\leq x\leq 2\\
&~~~~~~~~~~~~~~~~~~~~(\delta\mbox{-neighborhood of }x~\mbox{is }1~\mbox{or less.})\\
&\implies \left|x-1\right|<2\epsilon=\delta~~~~\mbox{(Because, }|x-1|\leq 1)\\
\end{align*}
$\therefore~\forall~\delta=\inf(1,2\epsilon),~\left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
