# For those that never learned well the epsilon-delta proofs

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• April 4th 2012, 07:00 PM
DKSanders123
Re: For those that never learnt well the epsilon-delta proofs
A great refresher. I'm actually working on those tonight, since in series and sequences the limits come back!! I was hoping they would disappear after the derivative ;)
• July 16th 2012, 02:05 AM
nhellie16
Re: For those that never learnt well the epsilon-delta proofs
thanks for the file.. it helps me lot... By the way its my birthday today.. thanks guys.. God bless..
• August 20th 2012, 08:57 AM
ebb
Re: For those that never learnt well the epsilon-delta proofs
Could I just check that my answer to this question is correct? Well, can I check that the d in my answer is correct at any rate!
Give an  e d  proof that the function
f(x) = (x^2 + 1)/(x + 1)
is continuous at x = 1.
I got d=min(1, 1.5e)
Thanks!
• August 20th 2012, 09:26 AM
ebb
Re: For those that never learnt well the epsilon-delta proofs
Also for f(x) = e^|x| + 1 at x=0. I got d=ln( epsilon + 1 )
• September 19th 2012, 04:23 AM
mittevans
Re: For those that never learnt well the epsilon-delta proofs
Delta-Epsilon proofs well-past my BC calc days when I was taking an introductory proofs class. I assume that many individuals going into or emerging of Calc just do not have evidences down sufficient yet to be able to whip one out.
• September 19th 2012, 05:32 AM
Prove It
Re: For those that never learnt well the epsilon-delta proofs
Quote:

Originally Posted by ebb
Could I just check that my answer to this question is correct? Well, can I check that the d in my answer is correct at any rate!
Give an e d proof that the function
f(x) = (x^2 + 1)/(x + 1)
is continuous at x = 1.
I got d=min(1, 1.5e)
Thanks!

To show \displaystyle \begin{align*} \frac{x^2 + 1}{x + 1} \end{align*} is continuous at \displaystyle \begin{align*} x = 1 \end{align*}, you need to show \displaystyle \begin{align*} \lim_{x \to 1}\frac{x^2 + 1}{x + 1} = 1 \end{align*}, i.e. that \displaystyle \begin{align*} |x - 1| < \delta \implies \left| \frac{x^2 + 1}{x + 1} - 1 \right| < \epsilon \end{align*}. Working on the second inequality we have

\displaystyle \begin{align*} \left| \frac{x^2 + 1}{x + 1} - 1 \right| &< \epsilon \\ \left| \frac{x^2 + 1}{x + 1} - \frac{x + 1}{x + 1} \right| &< \epsilon \\ \left| \frac{x^2 - x}{x + 1} \right| &< \epsilon \\ \left|x - 1\right| \left| \frac{x}{x + 1} \right| &< \epsilon \\ \left|x - 1\right| \left| 1 - \frac{1}{x + 1} \right| &< \epsilon \end{align*}

It would now help us if we had an upper bound on \displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| \end{align*}, to give us the maximum value of \displaystyle \begin{align*} \left| x - 1 \right| \left| 1 - \frac{1}{x + 1} \right| \end{align*}. If we bound \displaystyle \begin{align*} |x - 1| < 1 \end{align*} (say), that gives

\displaystyle \begin{align*} |x - 1| &< 1 \\ -1 < x - 1 &< 1 \\ 0 < x &< 2 \\ 1 < x + 1 &< 3 \\ \frac{1}{3} < \frac{1}{x + 1} &< 1 \\ \frac{1}{|x + 1|} &< 1 \end{align*}

and to get an upper bound on \displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| \end{align*} we can use the Triangle Inequality

\displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| &\leq |1| + \left| -\frac{1}{x + 1} \right| \\ &= 1 + \frac{1}{|x + 1|} \\ &\leq 1 + 1 \\ &= 2 \end{align*}

So that gives

\displaystyle \begin{align*} \left| x - 1 \right| \left| 1 - \frac{1}{x + 1} \right| &< \epsilon \\ 2|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{2} \end{align*}

Therefore choosing \displaystyle \begin{align*} \delta = \min\left\{ 1, \frac{\epsilon}{2} \right\} \end{align*} and reversing the process will finish the proof.
• September 19th 2012, 05:46 AM
Prove It
Re: For those that never learnt well the epsilon-delta proofs
Quote:

Originally Posted by ebb
Also for f(x) = e^|x| + 1 at x=0. I got d=ln( epsilon + 1 )

To show that \displaystyle \begin{align*} e^{|x|} + 1 \end{align*} is continuous at \displaystyle \begin{align*} x = 0 \end{align*}, you need to show \displaystyle \begin{align*} |x - 0| < \delta \implies \left| e^{|x|} + 1 - 2 \right| < \epsilon \end{align*}. Working on the second inequality we have

\displaystyle \begin{align*} \left| e^{|x|} - 1 \right| &< \epsilon \\ \left| e^{|x|} \right| - |1| &< \epsilon \textrm{ since } \left| e^{|x|} \right| - |1| \leq \left| e^{|x|} - 1 \right| \textrm{ by the Reverse Triangle Inequality } \\ e^{|x|} - 1 &< \epsilon \\ e^{|x|} &< \epsilon + 1 \\ |x| &< \ln{\left( \epsilon + 1 \right)} \end{align*}

So choosing \displaystyle \begin{align*} \delta = \ln{\left( \epsilon + 1 \right)} \end{align*} and reversing the process will complete the proof.
• October 13th 2012, 01:06 PM
Antiwasserstoff
Re: For those that never learnt well the epsilon-delta proofs
Hello! I hope its right to ask here, since i wanted to use a example from the pdf file, I am just using the proof of example 3 to show my problem.
I almost understand every single step to make the proof, the mainproblem is this step:
|x^2 - 25| = |x + 5||x - 5| < 11|x-5|

I know to get the 11|x-5|, but i dont get why it is less than my original function? In fact, its one of the first times i really work with inequallities, could someone explain me this little simple step? i solved |x+5| for 11, and thats why i can replace |x+5| with 11 which gives me 11|x-5|, but why is this new function < than my old function? arent both functions the same until i put my delta in?
• November 19th 2012, 12:44 AM
jesong
Re: For those that never learnt well the epsilon-delta proofs
I was taking an introductory proofs class. I think that most people going into or coming out of Calc just don't have proofs down enough yet to be able to whip one out.Luxury Goods Store
• November 19th 2012, 08:25 AM
Prove It
Re: For those that never learnt well the epsilon-delta proofs
Quote:

Originally Posted by Antiwasserstoff
Hello! I hope its right to ask here, since i wanted to use a example from the pdf file, I am just using the proof of example 3 to show my problem.
I almost understand every single step to make the proof, the mainproblem is this step:
|x^2 - 25| = |x + 5||x - 5| < 11|x-5|

I know to get the 11|x-5|, but i dont get why it is less than my original function? In fact, its one of the first times i really work with inequallities, could someone explain me this little simple step? i solved |x+5| for 11, and thats why i can replace |x+5| with 11 which gives me 11|x-5|, but why is this new function < than my old function? arent both functions the same until i put my delta in?

Because you are making \displaystyle \begin{align*} |x - 5| < 1 \end{align*} that means

\displaystyle \begin{align*} -1 < x - 5 &< 1 \\ 4 < x &< 6 \\ 9 < x + 5 &< 11 \\ |x + 5| &< 11 \\ |x + 5||x - 5| &< 11|x - 5| \end{align*}
• December 10th 2012, 12:43 AM
kingsure
Re: For those that never learnt well the epsilon-delta proofs
I understand that a little bit but I am having a hard time with how to start my proof.

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• September 27th 2013, 08:47 PM
mathlover10
Re: For those that never learnt well the epsilon-delta proofs
thanks for the nice paper! so in example 3 do you choose delta=1 arbitrarily and M is equal to the upperbound of 11? can you explain further what the meaning of the M derivation is?
for instance if you choose in ex 3 delta=0.1 instead of 1 then would get M=10.1 and define delta as min{0.1, 10.1} and is that an equivalent proof?
do most mathematicians learn this very quickly?
• September 28th 2013, 03:59 AM
Prove It
Re: For those that never learnt well the epsilon-delta proofs
The idea is that as long as your x is sufficiently close to the c value, then you are guaranteed for your function to be within your tolerance of $\displaystyle \epsilon$. So yes, the value $\displaystyle \delta = 1$ was chosen arbitrarily. Any number smaller than this will also work as we know this value works.
• January 9th 2014, 07:37 PM
JeanGunter
Re: For those that never learnt well the epsilon-delta proofs
Quote:

Originally Posted by matheagle
Student's eye glaze over when I mention this.
Plus books are dropping it, so don't blame your professors.
All books are becoming more and more watered down.
Some schools don't even teach trig substitution and I see hyperbolics are disappearing too...
Zill's DE book used to have lots of problems with cosh and sinh and now it hardly has any.
That stuff is really handy in DE.

Nice! I believe that! ;)
• March 19th 2014, 06:08 AM