show that the function $f(x)= frac{x^2+1}{x+1}$ is continuous at the point $x_0=1$












6












$begingroup$


Here is my attempt.



So starting with $f(x_0)=f(1)=1$ and $epsilon >0$, I want to find $delta >0$ such that if $|x-1|< delta$ I get $|f(x)-f(x_0)| leq epsilon$.



I started studying $|f(x)-f(x_0)|$ as follows



$|f(x)-f(x_0)|= |frac{x^2+1}{x+1}-1|= |frac{x^2-x}{x+1}|= |x frac{x-1}{x+1}| leq |frac{x}{x+1}| delta$



But now I cannot find a way to conclude. I want to find an upper bound for $|frac{x}{x+1}|$ and then find an expression for $delta$ in terms of $epsilon$



$|f(x)-f(x_0)|= |frac{x}{x+1}| delta leq frac{(delta +1)delta}{2- delta}$



From now on I tried to set $ frac{(delta +1)delta}{2- delta} = epsilon$ and solve for $delta$ but this didn't work.



I then noticed that $frac{(delta +1)delta}{2- delta} leq frac{(delta +1)^2}{2- delta}$ from where I tried solving $frac{(delta +1)^2}{2- delta}=epsilon$ but again, I failed to find an applicable solution.



I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.



Thank you for your help.






real-analysis continuity epsilon-delta







share|cite|improve this question









$endgroup$












  • $begingroup$
    try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
    $endgroup$
    – Jagol95
    Dec 16 '18 at 20:32
















6












$begingroup$


Here is my attempt.



So starting with $f(x_0)=f(1)=1$ and $epsilon >0$, I want to find $delta >0$ such that if $|x-1|< delta$ I get $|f(x)-f(x_0)| leq epsilon$.



I started studying $|f(x)-f(x_0)|$ as follows



$|f(x)-f(x_0)|= |frac{x^2+1}{x+1}-1|= |frac{x^2-x}{x+1}|= |x frac{x-1}{x+1}| leq |frac{x}{x+1}| delta$



But now I cannot find a way to conclude. I want to find an upper bound for $|frac{x}{x+1}|$ and then find an expression for $delta$ in terms of $epsilon$



$|f(x)-f(x_0)|= |frac{x}{x+1}| delta leq frac{(delta +1)delta}{2- delta}$



From now on I tried to set $ frac{(delta +1)delta}{2- delta} = epsilon$ and solve for $delta$ but this didn't work.



I then noticed that $frac{(delta +1)delta}{2- delta} leq frac{(delta +1)^2}{2- delta}$ from where I tried solving $frac{(delta +1)^2}{2- delta}=epsilon$ but again, I failed to find an applicable solution.



I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.



Thank you for your help.






real-analysis continuity epsilon-delta







share|cite|improve this question









$endgroup$












  • $begingroup$
    try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
    $endgroup$
    – Jagol95
    Dec 16 '18 at 20:32














6












6








6





$begingroup$


Here is my attempt.



So starting with $f(x_0)=f(1)=1$ and $epsilon >0$, I want to find $delta >0$ such that if $|x-1|< delta$ I get $|f(x)-f(x_0)| leq epsilon$.



I started studying $|f(x)-f(x_0)|$ as follows



$|f(x)-f(x_0)|= |frac{x^2+1}{x+1}-1|= |frac{x^2-x}{x+1}|= |x frac{x-1}{x+1}| leq |frac{x}{x+1}| delta$



But now I cannot find a way to conclude. I want to find an upper bound for $|frac{x}{x+1}|$ and then find an expression for $delta$ in terms of $epsilon$



$|f(x)-f(x_0)|= |frac{x}{x+1}| delta leq frac{(delta +1)delta}{2- delta}$



From now on I tried to set $ frac{(delta +1)delta}{2- delta} = epsilon$ and solve for $delta$ but this didn't work.



I then noticed that $frac{(delta +1)delta}{2- delta} leq frac{(delta +1)^2}{2- delta}$ from where I tried solving $frac{(delta +1)^2}{2- delta}=epsilon$ but again, I failed to find an applicable solution.



I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.



Thank you for your help.






real-analysis continuity epsilon-delta







share|cite|improve this question









$endgroup$




Here is my attempt.



So starting with $f(x_0)=f(1)=1$ and $epsilon >0$, I want to find $delta >0$ such that if $|x-1|< delta$ I get $|f(x)-f(x_0)| leq epsilon$.



I started studying $|f(x)-f(x_0)|$ as follows



$|f(x)-f(x_0)|= |frac{x^2+1}{x+1}-1|= |frac{x^2-x}{x+1}|= |x frac{x-1}{x+1}| leq |frac{x}{x+1}| delta$



But now I cannot find a way to conclude. I want to find an upper bound for $|frac{x}{x+1}|$ and then find an expression for $delta$ in terms of $epsilon$



$|f(x)-f(x_0)|= |frac{x}{x+1}| delta leq frac{(delta +1)delta}{2- delta}$



From now on I tried to set $ frac{(delta +1)delta}{2- delta} = epsilon$ and solve for $delta$ but this didn't work.



I then noticed that $frac{(delta +1)delta}{2- delta} leq frac{(delta +1)^2}{2- delta}$ from where I tried solving $frac{(delta +1)^2}{2- delta}=epsilon$ but again, I failed to find an applicable solution.



I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.



Thank you for your help.






real-analysis continuity epsilon-delta




real-analysis continuity epsilon-delta






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 20:03









Alain.KlbtrAlain.Klbtr

927




927












  • $begingroup$
    try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
    $endgroup$
    – Jagol95
    Dec 16 '18 at 20:32


















  • $begingroup$
    try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
    $endgroup$
    – Jagol95
    Dec 16 '18 at 20:32
















$begingroup$
try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
$endgroup$
– Jagol95
Dec 16 '18 at 20:32




$begingroup$
try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell.
$endgroup$
– Jagol95
Dec 16 '18 at 20:32










2 Answers
2






active

oldest

votes


















3












$begingroup$

Unfortunately, $$frac{x}{x+1}=1-frac1{x+1}$$ is an unbounded function, so that $leftlvertfrac{x}{x+1}rightrvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.



So long as $x$ is positive, $frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?



Consequently, picking some arbitrary $alphain(0,1),$ we need only make sure that $delta=min{alpha,epsilon},$ at which point we'll have $$leftlvertfrac{x}{x+1}rightrvertdelta<deltaleepsilon$$ whenever $|x-1|<delta,$ which completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:51








  • 1




    $begingroup$
    That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:52










  • $begingroup$
    Many thanks for your kind help
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53










  • $begingroup$
    +1: Well and helpfully explained. :-)
    $endgroup$
    – E-mu
    Dec 16 '18 at 23:53





















2












$begingroup$

You could choose a $delta$ which ensures both $|frac{x}{x+1}| < 1$ and $delta < epsilon$ so you can easily extend the steps in your first line of workings.



It can be shown that $|frac{x}{x+1}| < 1 text{if and only if} x > -frac 12$, and we know $ 0 < |x-1| < delta iff 1 - delta < x < 1 + delta$ and $x neq 1$. Therefore, we want a $delta$ such that $1 - delta geq -frac 12 iff delta leq frac 32$.



We have narrowed down our requirements for $delta$ to




  • $delta leq frac 32$

  • $delta < epsilon$

  • $delta > 0$


I would then suggest finding a function of a real positive variable which satisfies these conditions.



An example of such $delta$ is $delta = frac{epsilon}{epsilon+1}$.



$textbf{Edit}$: It is fine to relax the condition $delta < epsilon$ to $delta leq epsilon$ as shown in Cameron's answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1: Very elegant! I like it.
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:49












  • $begingroup$
    Thank you ! That's a very clever way of doing it.
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Unfortunately, $$frac{x}{x+1}=1-frac1{x+1}$$ is an unbounded function, so that $leftlvertfrac{x}{x+1}rightrvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.



So long as $x$ is positive, $frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?



Consequently, picking some arbitrary $alphain(0,1),$ we need only make sure that $delta=min{alpha,epsilon},$ at which point we'll have $$leftlvertfrac{x}{x+1}rightrvertdelta<deltaleepsilon$$ whenever $|x-1|<delta,$ which completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:51








  • 1




    $begingroup$
    That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:52










  • $begingroup$
    Many thanks for your kind help
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53










  • $begingroup$
    +1: Well and helpfully explained. :-)
    $endgroup$
    – E-mu
    Dec 16 '18 at 23:53


















3












$begingroup$

Unfortunately, $$frac{x}{x+1}=1-frac1{x+1}$$ is an unbounded function, so that $leftlvertfrac{x}{x+1}rightrvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.



So long as $x$ is positive, $frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?



Consequently, picking some arbitrary $alphain(0,1),$ we need only make sure that $delta=min{alpha,epsilon},$ at which point we'll have $$leftlvertfrac{x}{x+1}rightrvertdelta<deltaleepsilon$$ whenever $|x-1|<delta,$ which completes the proof.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:51








  • 1




    $begingroup$
    That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:52










  • $begingroup$
    Many thanks for your kind help
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53










  • $begingroup$
    +1: Well and helpfully explained. :-)
    $endgroup$
    – E-mu
    Dec 16 '18 at 23:53
















3












3








3





$begingroup$

Unfortunately, $$frac{x}{x+1}=1-frac1{x+1}$$ is an unbounded function, so that $leftlvertfrac{x}{x+1}rightrvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.



So long as $x$ is positive, $frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?



Consequently, picking some arbitrary $alphain(0,1),$ we need only make sure that $delta=min{alpha,epsilon},$ at which point we'll have $$leftlvertfrac{x}{x+1}rightrvertdelta<deltaleepsilon$$ whenever $|x-1|<delta,$ which completes the proof.






share|cite|improve this answer









$endgroup$



Unfortunately, $$frac{x}{x+1}=1-frac1{x+1}$$ is an unbounded function, so that $leftlvertfrac{x}{x+1}rightrvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.



So long as $x$ is positive, $frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?



Consequently, picking some arbitrary $alphain(0,1),$ we need only make sure that $delta=min{alpha,epsilon},$ at which point we'll have $$leftlvertfrac{x}{x+1}rightrvertdelta<deltaleepsilon$$ whenever $|x-1|<delta,$ which completes the proof.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 21:18









Cameron BuieCameron Buie

85.3k772158




85.3k772158












  • $begingroup$
    thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:51








  • 1




    $begingroup$
    That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:52










  • $begingroup$
    Many thanks for your kind help
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53










  • $begingroup$
    +1: Well and helpfully explained. :-)
    $endgroup$
    – E-mu
    Dec 16 '18 at 23:53




















  • $begingroup$
    thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:51








  • 1




    $begingroup$
    That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:52










  • $begingroup$
    Many thanks for your kind help
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53










  • $begingroup$
    +1: Well and helpfully explained. :-)
    $endgroup$
    – E-mu
    Dec 16 '18 at 23:53


















$begingroup$
thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:51






$begingroup$
thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $delta = min${$epsilon;frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $delta$ because $|frac{x}{x+1}| leq frac{delta +1}{2- delta}$ and so I tried to solve $frac{delta +1}{2- delta} leq 1$ and found that $delta$ should be in $(- infty ; frac{1}{2}] bigcup (2; infty)$
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:51






1




1




$begingroup$
That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
$endgroup$
– Cameron Buie
Dec 16 '18 at 21:52




$begingroup$
That absolutely works. Any $alphain(0,1)$ will do the job, including $alpha=frac12.$
$endgroup$
– Cameron Buie
Dec 16 '18 at 21:52












$begingroup$
Many thanks for your kind help
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:53




$begingroup$
Many thanks for your kind help
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:53












$begingroup$
+1: Well and helpfully explained. :-)
$endgroup$
– E-mu
Dec 16 '18 at 23:53






$begingroup$
+1: Well and helpfully explained. :-)
$endgroup$
– E-mu
Dec 16 '18 at 23:53













2












$begingroup$

You could choose a $delta$ which ensures both $|frac{x}{x+1}| < 1$ and $delta < epsilon$ so you can easily extend the steps in your first line of workings.



It can be shown that $|frac{x}{x+1}| < 1 text{if and only if} x > -frac 12$, and we know $ 0 < |x-1| < delta iff 1 - delta < x < 1 + delta$ and $x neq 1$. Therefore, we want a $delta$ such that $1 - delta geq -frac 12 iff delta leq frac 32$.



We have narrowed down our requirements for $delta$ to




  • $delta leq frac 32$

  • $delta < epsilon$

  • $delta > 0$


I would then suggest finding a function of a real positive variable which satisfies these conditions.



An example of such $delta$ is $delta = frac{epsilon}{epsilon+1}$.



$textbf{Edit}$: It is fine to relax the condition $delta < epsilon$ to $delta leq epsilon$ as shown in Cameron's answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1: Very elegant! I like it.
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:49












  • $begingroup$
    Thank you ! That's a very clever way of doing it.
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53
















2












$begingroup$

You could choose a $delta$ which ensures both $|frac{x}{x+1}| < 1$ and $delta < epsilon$ so you can easily extend the steps in your first line of workings.



It can be shown that $|frac{x}{x+1}| < 1 text{if and only if} x > -frac 12$, and we know $ 0 < |x-1| < delta iff 1 - delta < x < 1 + delta$ and $x neq 1$. Therefore, we want a $delta$ such that $1 - delta geq -frac 12 iff delta leq frac 32$.



We have narrowed down our requirements for $delta$ to




  • $delta leq frac 32$

  • $delta < epsilon$

  • $delta > 0$


I would then suggest finding a function of a real positive variable which satisfies these conditions.



An example of such $delta$ is $delta = frac{epsilon}{epsilon+1}$.



$textbf{Edit}$: It is fine to relax the condition $delta < epsilon$ to $delta leq epsilon$ as shown in Cameron's answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1: Very elegant! I like it.
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:49












  • $begingroup$
    Thank you ! That's a very clever way of doing it.
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53














2












2








2





$begingroup$

You could choose a $delta$ which ensures both $|frac{x}{x+1}| < 1$ and $delta < epsilon$ so you can easily extend the steps in your first line of workings.



It can be shown that $|frac{x}{x+1}| < 1 text{if and only if} x > -frac 12$, and we know $ 0 < |x-1| < delta iff 1 - delta < x < 1 + delta$ and $x neq 1$. Therefore, we want a $delta$ such that $1 - delta geq -frac 12 iff delta leq frac 32$.



We have narrowed down our requirements for $delta$ to




  • $delta leq frac 32$

  • $delta < epsilon$

  • $delta > 0$


I would then suggest finding a function of a real positive variable which satisfies these conditions.



An example of such $delta$ is $delta = frac{epsilon}{epsilon+1}$.



$textbf{Edit}$: It is fine to relax the condition $delta < epsilon$ to $delta leq epsilon$ as shown in Cameron's answer.






share|cite|improve this answer











$endgroup$



You could choose a $delta$ which ensures both $|frac{x}{x+1}| < 1$ and $delta < epsilon$ so you can easily extend the steps in your first line of workings.



It can be shown that $|frac{x}{x+1}| < 1 text{if and only if} x > -frac 12$, and we know $ 0 < |x-1| < delta iff 1 - delta < x < 1 + delta$ and $x neq 1$. Therefore, we want a $delta$ such that $1 - delta geq -frac 12 iff delta leq frac 32$.



We have narrowed down our requirements for $delta$ to




  • $delta leq frac 32$

  • $delta < epsilon$

  • $delta > 0$


I would then suggest finding a function of a real positive variable which satisfies these conditions.



An example of such $delta$ is $delta = frac{epsilon}{epsilon+1}$.



$textbf{Edit}$: It is fine to relax the condition $delta < epsilon$ to $delta leq epsilon$ as shown in Cameron's answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 0:22

























answered Dec 16 '18 at 21:38









E-muE-mu

787417




787417












  • $begingroup$
    +1: Very elegant! I like it.
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:49












  • $begingroup$
    Thank you ! That's a very clever way of doing it.
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53


















  • $begingroup$
    +1: Very elegant! I like it.
    $endgroup$
    – Cameron Buie
    Dec 16 '18 at 21:49












  • $begingroup$
    Thank you ! That's a very clever way of doing it.
    $endgroup$
    – Alain.Klbtr
    Dec 16 '18 at 21:53
















$begingroup$
+1: Very elegant! I like it.
$endgroup$
– Cameron Buie
Dec 16 '18 at 21:49






$begingroup$
+1: Very elegant! I like it.
$endgroup$
– Cameron Buie
Dec 16 '18 at 21:49














$begingroup$
Thank you ! That's a very clever way of doing it.
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:53




$begingroup$
Thank you ! That's a very clever way of doing it.
$endgroup$
– Alain.Klbtr
Dec 16 '18 at 21:53


















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