<s>[INST] Proving limits of functions using first principles Prove using first principles that $\lim_{x \to 2}$ ($\frac{x}{1+x}$) = $\frac{2}{3}$
I know that you need to use a $\delta$-$\varepsilon$ proof where you fix $\varepsilon > 0$ and find $\delta > 0$ such that $0<|x - 2|<\delta$ $\implies$ $|\frac{x}{1+x}$ - $\frac{2}{3}|$ < $\varepsilon$
I got $|x-2| < 3\varepsilon|1+x|$ from $|\frac{x}{1+x} - \frac{2}{3}| < \varepsilon$. I know how you finish off the proof once you have found a $\delta$. The part I am not quite sure about is if I can use $\delta$ = 3$\varepsilon$|1+x| directly or if I need to get rid of the x somehow
 [/INST] You need to "get rid of the $x$" by solving the inequality. Start fixing $\epsilon>0$ and consider
$$
\left|\frac{x}{1+x}-\frac{2}{3}\right|<\epsilon.
$$
As soon as $0<\epsilon<\frac 13$, by solving the previous inequality we get $\frac{2-3\epsilon}{1+3\epsilon}<x<\frac{2+3\epsilon}{1-3\epsilon}$. We would like to find $\delta>0$ such that this interval in $x$ is contained in $(2-\delta,2+\delta)$. Note that 
$$
\frac{2-3\epsilon}{1+3\epsilon}=2-\frac{9\epsilon}{1+3\epsilon}\quad\text{and}\quad \frac{2+3\epsilon}{1-3\epsilon}=2+\frac{9\epsilon}{1-3\epsilon}.
$$ 
Since $\frac{9\epsilon}{1+3\epsilon}<\frac{9\epsilon}{1-3\epsilon}$, we can choose $\delta=\frac{9\epsilon}{1+3\epsilon}$ and the definition of limit is satisfied.
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