My Math Forum show (x+2)/(x^2-x) is continuous (epsilon-delta)

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 December 22nd, 2011, 07:17 PM #1 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 show (x+2)/(x^2-x) is continuous (epsilon-delta) Ok, I've been playing with the function $f(x)=\frac{x+2}{x^2-x}$ and I'm just not having any luck showing that it is continuous on the open interval (0,1). Any hints? Is there a trick I'm not seeing, or do I just need to make some estimates to make it work somehow? I want to show it's continuous using the epsilon-delta defn, so I've been playing with that, but not having much luck. thanks!
 December 22nd, 2011, 07:38 PM #2 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: show (x+2)/(x^2-x) is continuous (epsilon-delta) So x and (x-1) are continuous on (0,1), so x(x-1) also. Furthermore, if you denote g(x)=x(x-1), this function is such that $g(x)\neq 0$ on (0,1), thus $\frac{1}{g(x)}$ is continuous (theorem). Finally x+é i scontinuous, product of continuous functions is continuous, so (x+2)(x^2-x) is continuous...Concerning the other problems of uniform convergence, I have to say I don't have any solution! (I would have to spend time on it and I can not now)
 December 22nd, 2011, 07:40 PM #3 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Re: show (x+2)/(x^2-x) is continuous (epsilon-delta) Ok would the following work? Let $\varepsilon>0$ be given and let $\delta=\varepsilon\cdot \left|\frac{(x^2-x)(y^2-y)}{xy+2y+2x-2}\right|$ (from now on, put $M_{xy}:=\left|\frac{(x^2-x)(y^2-y)}{xy+2y+2x-2}\right|$). Then whenever $|x-y|<\delta$, we get $|f(x)-f(y)|=|x-y|\frac{1}{M_{xy}}<\delta\frac{1}{M_{xy}}=\varepsi lon$. Does this work to let delta depend on x and y like that or am I missing something else here? Thanks.
December 22nd, 2011, 07:42 PM   #4
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Re: show (x+2)/(x^2-x) is continuous (epsilon-delta)

Quote:
 Originally Posted by Dougy So x and (x-1) are continuous on (0,1), so x(x-1) also. Furthermore, if you denote g(x)=x(x-1), this function is such that $g(x)\neq 0$ on (0,1), thus $\frac{1}{g(x)}$ is continuous (theorem). Finally x+é i scontinuous, product of continuous functions is continuous, so (x+2)(x^2-x) is continuous...Concerning the other problems of uniform convergence, I have to say I don't have any solution! (I would have to spend time on it and I can not now)

Thanks, but I am actually looking for a way to show continuity using the definition of epsilon-dela, even though what you stated here is perfectly good and much more straightforward! But I'd specifically like an epsilon-delta proof..perhaps you could take a look at my last post? (I didn't put in all the details, but hopefully I did the computations correctly).

December 22nd, 2011, 10:00 PM   #5
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Re: show (x+2)/(x^2-x) is continuous (epsilon-delta)

Quote:
 Originally Posted by laurence Ok would the following work? Let $\varepsilon>0$ be given and let $\delta=\varepsilon\cdot \left|\frac{(x^2-x)(y^2-y)}{xy+2y+2x-2}\right|$ (from now on, put $M_{xy}:=\left|\frac{(x^2-x)(y^2-y)}{xy+2y+2x-2}\right|$). Then whenever $|x-y|<\delta$, we get $|f(x)-f(y)|=|x-y|\frac{1}{M_{xy}}<\delta\frac{1}{M_{xy}}=\varepsi lon$. Does this work to let delta depend on x and y like that or am I missing something else here? Thanks.

say we choose "a" to be the continuous point:

|f(x)-f(a)|< epsilon whenever |x-a| < delta

computing this, we get:

${\frac {x+2}{{x}^{2}-x}}-{\frac {a+2}{{a}^{2}-a}}$

turning this into one fraction gives us:

${\frac {x{a}^{2}+2\,{a}^{2}-2\,a-{x}^{2}a-2\,{x}^{2}+2\,x}{ \left( {x}^{2}-x \right) \left( {a}^{2}-a \right) }}$

Take only the numerator of this, and do long division by ${x}-{a}$

Which gives us $-xa-2\,a+2-2\,x$

$|x-a|<\delta\rightarrow|f(x)-f(a)<\varepsilon$

letting $\delta=\varepsilon*({\frac { \left( {x}^{2}-x \right) \left( {a}^{2}-a \right) }{-xa-2\,a+2-2\,x}})$

which then gives:

$|x-a|=<\delta=\varepsilon*({\frac { \left( {x}^{2}-x \right) \left( {a}^{2}-a \right) }{-xa-2\,a+2-2\,x}})$

Now we choose on the modulus (absolute value), if you take the modulus to be multiplied by negative 1, you have your equation (-xa-2a+2-2x), if we stick with positive 1, you just your equation multiplied by -1 aka the equation I have derived:

$|{\frac { \left( x-a \right) \left( -xa-2\,a+2-2\,x \right) }{ \left( {x}^{2}-x \right) \left( {a}^{2}-a \right) }}|=<\varepsilon= |f(x)-f(a)|<\varepsilon=$

And thus satisfying the equation, with some more work you can state it rigorously.

With proof of continuous functions and general proof of limits, the more absurd they get (cubes and powers of 4,5,6, etc), often the composition of functions, as Dougie has outlined, is the best way to approach solving the equation.

 December 23rd, 2011, 06:48 AM #6 Senior Member   Joined: Apr 2010 Posts: 451 Thanks: 1 Re: show (x+2)/(x^2-x) is continuous (epsilon-delta) First of all ,to avoid any mistakes ,do we all agree that: $|(a^2-a)(x+2)-(a+2)(x^2-x)|\leq |a-x||ax+2(x+a)-2|$
 December 24th, 2011, 04:52 AM #7 Senior Member   Joined: Apr 2010 Posts: 451 Thanks: 1 Re: show (x+2)/(x^2-x) is continuous (epsilon-delta) Given ?>? ,choose :$\delta=$min $(\frac{|a^2-a|}{6},\frac{\epsilon(a^2-a)^2}{14})$ Hence if............ |x-a|
January 5th, 2012, 08:43 PM   #8
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Re: show (x+2)/(x^2-x) is continuous (epsilon-delta)

Quote:
 Originally Posted by outsos Given ?>? ,choose :$\delta=$min $(\frac{|a^2-a|}{6},\frac{\epsilon(a^2-a)^2}{14})$ Hence if............ |x-a|

Could you possibly elaborate on how you found your delta? I had something quite different, but it's very possible we just used different estimations i guess.. if we are showing the function is continuous at $c$, I had that picking $\delta=\min\{\frac{c}{2},\frac{1-c}{2},\frac{c^2(1-c)^2}{4}\varepsilon\}$ should work.

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### USING EPSILON DELTA TO PROVE THAT PRODUCTS OF CONTINUOUS FUNCTION IS CONTINUOUS

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