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 $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
 A: What you did cannot possibly work. Since, for any $x\in\mathbb R$,$$\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13\geqslant\frac13,$$ if $\varepsilon\in\left(0,\frac13\right)$, then there is no $\delta>0$ such that$$\lvert x-1\rvert<\delta\implies\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13<\varepsilon.$$Note that\begin{align}\left\lvert\frac1{2+\sqrt x}-\frac13\right\rvert&=\left\lvert\frac{1-\sqrt x}{3\left(2+\sqrt x\right)}\right\rvert\\&\leqslant\frac{\left\lvert\sqrt x-1\right\rvert}6\\&=\frac{\left\lvert\left(\sqrt x-1\right)\left(\sqrt x+1\right)\right\rvert}{6\left(\sqrt x+1\right)}\\&\leqslant\frac{\left\lvert x-1\right\rvert}6.\end{align}So, for each $\varepsilon>0$, take $\delta=6\varepsilon$.
A: You need to relate $\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right|$ with $|x-1|$ in a proper way:
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| = \left| \frac{1 - \sqrt{x}}{3(2+\sqrt{x})} \right| = \left| \frac{1-x}{3(1+\sqrt{x})(2+\sqrt{x})} \right| \leq \frac{|x-1|}{3\cdot1\cdot2} = \frac{|x-1|}{6}$ so for a given 
$\epsilon > 0$, choosing $\delta=\epsilon$ gives you for $0<|x-1|<\delta = \epsilon$, 
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| \leq \frac{|x-1|}{6} < \frac{\delta}{6} = \frac{\epsilon}{6} < \epsilon$.
A: 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 $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
