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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.

Luckyian
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    It is difficult to find the question you are asking. Any chance you can separate that part out a bit so it's easier to see? – abiessu Jan 09 '24 at 23:52
  • Please check it now, the problem that I was facing is, when I choose delta to be 2e and then to prove the implication – Luckyian Jan 10 '24 at 00:10
  • Not sure about the inequality $\frac {|x-5|}{|x-3|} < \frac{1}{3}$ for the $\delta=1$ case. For $x=4.1$ (to satisfy $|x-5|<1$), the LHS $\frac {|x-5|}{|x-3|} \approx 0.82 $. – peterwhy Jan 10 '24 at 00:19
  • $|x-5| < 1$ ==> $\frac{|x-5|}{|x-3|} < \frac{1}{|x-3|}$ now the left side of this will get the minimum when |x-3| gets max in right hand side denominator, so i interchanged it with 3, then it was $\frac{|x-5|}{|x-3|} < \frac{1}{3}$ now I just multiplied both sides with 1/2, so that I will get 2 in denominator for the left hand side. Then it will be $\frac{|x-5|}{2|x-3|} < \frac{1}{6}$ , now as the delta as 1 which means $ 1< 2\epsilon $, I can rewrite it as, $\frac{1}{2} < \epsilon $, so this how I used it, $|\frac{|-x+5}{2(x-3)} |< \frac{1}{6} < \frac{1}{2} < \epsilon $ – Luckyian Jan 10 '24 at 00:39
  • But then for $x=4.1$, $\frac {|x-5|}{|x-3|} = \frac{.9}{1.1} = \frac{27}{11\times3} > \frac13$. – peterwhy Jan 10 '24 at 00:45
  • Then where I was going wrong :( – Luckyian Jan 10 '24 at 01:34
  • You are right that $\frac{1}{|x-3|}$ achieves it's minimum when $|x-3|$ and that under the constraint $|x-5| < 1$ the maximum of $|x-3|$ is $3$. But this actually means that $\frac{1}{|x-3|}$ is always at least as big as $\frac{1}{3}$ not smaller. That is $\frac{1}{|x-3|} > \frac{1}{3}$. To get the correct inequality you would have to find the maximum of $\frac{1}{|x-3|}$, which is 1 under the constraint $|x-5| < 1$. – NaturalLogZ Jan 10 '24 at 02:26
  • @NaturalLogZ ok so i replace it with just a 1 and both sides divide with 2, then it will be, $|\frac{x-5}{2(x-3)}| < \frac{1}{2} < \epsilon$ – Luckyian Jan 10 '24 at 04:30

1 Answers1

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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,

$$\lim_{x\to 5}\frac1{x-3} = \frac12.$$

peterwhy
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