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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 $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$

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    $\begingroup$ When you apply the triangle inequality on line 1 of your argument, you have lost the game: if $0 < \epsilon < 1/3$, you cannot make $|\ldots| + |1/3| < \epsilon$. $\endgroup$ – Rob Arthan Mar 11 '20 at 22:57
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    $\begingroup$ Your proof doesn't work. If $\varepsilon$ is small, then $\delta \lt 0$. $\endgroup$ – Robert Shore Mar 11 '20 at 22:58
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    $\begingroup$ A more fundamental problem (in the approach in your question as well as in last comment) is that you replace the inequality $|f(x) - L|<\epsilon$ with say $g(x) <\epsilon $ in order to do some simplification but you lose the chain of implication. Your new inequality should always be such that $g(x) <\epsilon\implies |f(x) - L|<\epsilon $ and not the other way round $|f(x) - L|<\epsilon \implies g(x) <\epsilon $. The original inequality as per definition is a target, the final in a chain of logical implications. $\endgroup$ – Paramanand Singh Mar 14 '20 at 19:26
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    $\begingroup$ You can think of the target inequality as a goal to be achieved and the intermediate inequality as one of ways to achieve that goal. You can't change the goalpost itself. $\endgroup$ – Paramanand Singh Mar 14 '20 at 19:28
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    $\begingroup$ Your answer below has the problem of changing the goalpost. Your new inequality does not imply old one. $\endgroup$ – Paramanand Singh Mar 14 '20 at 19:51
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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 $\delta=6\varepsilon$.

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  • $\begingroup$ Just Re-read the definition and I've been missing the detail "given any $\epsilon$>0$, there exists..." hence I must be able to get \epsilon arbitrarily close to zero as well. Thank you all for your assistance. $\endgroup$ – C-Web Mar 12 '20 at 1:19
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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 $0<|x-1|<\delta = \epsilon$,

$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| \leq \frac{|x-1|}{6} < \frac{\delta}{6} = \frac{\epsilon}{6} < \epsilon$.

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  • $\begingroup$ That's the exact approach my professor took. I agree that this is cleaner, but I'm trying to understand what in my thinking (or understanding of delta - epsilon) is flawed. $\endgroup$ – C-Web Mar 12 '20 at 1:01
  • $\begingroup$ Just Re-read the definition and I've been missing the detail "given any $\epsilon$>0$, there exists..." hence I must be able to get \epsilon arbitrarily close to zero as well. Thank you all for your assistance. $\endgroup$ – C-Web Mar 12 '20 at 1:19
  • $\begingroup$ I am happy to hear that you comprehend the definition of limit and meaning of "for all $\epsilon>0$" so well. You are welcome. $\endgroup$ – Can Turkun Mar 12 '20 at 1:42
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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 $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$

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    $\begingroup$ Are you really sure that if $0<|x-1|<\delta=\min(1,2\epsilon)\implies |1/(2+\sqrt{x})|<\epsilon$. $\endgroup$ – Paramanand Singh Mar 15 '20 at 15:48
  • $\begingroup$ @ParamanandSingh I'm inclined to say yes here.If $2\epsilon > 1$, then $\epsilon >\frac{1}{2}$ and $-\frac{1}{3}<f(x)-L<\frac{1}{2}$ holds. If $2\epsilon<1$, then $\epsilon<1/2$ and since $-2\epsilon<|x-1|<2\epsilon$ then $1-2\epsilon<x<1+2\epsilon$ ~~~OOOHHH! I see what you mean!.~~~~ We used the triangle inequality a good bit in complex analysis. Is it generally a mistake to use it in real variable analysis? Or, is it just one of those situational recognition things where sometimes it fits and sometimes it doesn't? $\endgroup$ – C-Web Mar 15 '20 at 17:11
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    $\begingroup$ the problem is not the triangle inequality. It is just a tool and you need to understand how to use a tool to achieve your goal. Sometime you may not a specific tool at all. Start by analyzing the difference $|f(x) - L|=\dfrac{|\sqrt{x} - 1|}{3(2+\sqrt{x})}$ and see if we can make it small by assuming that $|x-1|$ small. Usage of inequalities in analysis proofs in more about understanding them in words rather symbol manipulation. $\endgroup$ – Paramanand Singh Mar 16 '20 at 1:01
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    $\begingroup$ Thus for example one can write the above expression as $\dfrac{|x-1|}{3(1+\sqrt{x})(2+\sqrt{x})}$. And now things are much clearer. If one assumes $|x-1|$ small then the numerator becomes small and this ensures that the overall fraction is small provided we ensure denominator does not become too small. So the problem shifts now to analysis of denominator. And we have to ensure that it does not become too small. But this part is obvious as denominator anyway exceeds $3\cdot 2\cdot 1=6$. You can now understand why $\delta=6\epsilon$ will work as expected. $\endgroup$ – Paramanand Singh Mar 16 '20 at 1:06

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