<s>[INST] 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 $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
 [/INST] 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 $0<|x-1|<\delta = \epsilon$, 
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| \leq \frac{|x-1|}{6} < \frac{\delta}{6} = \frac{\epsilon}{6} < \epsilon$.
</s>