I was trying to prove $$\lim_{x\to5}\frac{1}{x-3}= \frac{1}{2}$$ i ended up by having these inequalities, $ -1 < |x-3| < 3$
$|x-5|<1$
$|x-5|<2\epsilon$
so now i have $\delta =$min{$1,2\epsilon$}, from here on i tried to prove $|x-5|< \delta$==>$|f(x) -L| < \epsilon$
let $\delta=1$, which means $1 < 2\epsilon$,i can rewrite it as $\frac{1}{2} < \epsilon$,
$|x-5| < 1$ ==> $\frac {|x-5|}{|x-3|} < \frac{1}{3}$ , i have written $3$ because whenever $|x-3|$ is at max, the minimum the right hand side will get, then multiplied both side with $\frac{1}{2}$, then it will be $\frac{1}{2}\frac{|x-5|}{|x-3|}<\frac{1}{6}$, now i can convert the left side into $|f(x) - L|$ form, coming to the right hand side this is how i concluded it, as from the above $\frac{1}{2} < \epsilon$ so, $\frac{1}{6} < \frac{1}{2} < \epsilon$ so replaced the left hand side with $\epsilon$.
now let $\delta = 2\epsilon$, so $2\epsilon < 1$
$|x-5| < 2\epsilon$ ==> $\frac {|x-5|}{|x-3|} < \frac{2\epsilon}{|x-3|}$, now what should place in of $|x-3|$ in the right hand side, i was thinking the left hand side gets minimum whenever $|x-3|$ was at its max should i place $3$, if its $3$ then how to conclude it will be $\epsilon$.
Thanks in advance.
For given $\epsilon >0$, pick
$$\delta = \dfrac{4\epsilon}{1+2\epsilon} >0.$$
Then for all $x$ satisfying $\left\lvert x-5\right\rvert <\delta$,
$$\begin{array}{rcl} -\delta <& x-5 &< \delta\\ 2-\delta <& x-3 &< 2+\delta\\ \dfrac{2+4\epsilon-4\epsilon}{1+2\epsilon } < & x-3 &< \dfrac{2+4\epsilon+4\epsilon}{1+2\epsilon }\\ \dfrac{2}{1+2\epsilon } < & x-3 &< \dfrac{2+8\epsilon}{1+2\epsilon }\\ \dfrac{1+2\epsilon }{2} > & \dfrac1{x-3} &> \dfrac{1+2\epsilon}{2+8\epsilon }\\ \dfrac{1}{2} +\epsilon > & \dfrac1{x-3} &> \dfrac12 - \epsilon + \dfrac{8\epsilon^2}{2+8\epsilon }\\ \epsilon > & \dfrac1{x-3} - \dfrac12 &> - \epsilon + \dfrac{8\epsilon^2}{2+8\epsilon } > -\epsilon\\ & \left\lvert\dfrac1{x-3}-\dfrac12\right\rvert &<\epsilon\\ \end{array}$$
By the epsilon-delta definition of limits,
$$\lim_{x\to 5}\frac1{x-3} = \frac12.$$