Use limit definition to prove

[John O' Meara]

Ues the limit definition to prove that the stated limit is correct. \lim_{x-&gt;-2} \frac{1}{x+1}=-1. The limit def' is |f(x)-L|<epsilon if 0<|x-a|< delta. So we have |\frac{1}{x+1} + 1| &lt; \epsilon if 0 &lt; |x- (-2)| &lt; \delta \mbox{ therefore } |\frac{1}{x+1}||x+2| &lt; \epslion \mbox{ if } 0 &lt; |x + 2| &lt; \delta. To bound \frac{1}{x+1} \mbox { let } -1 &lt; \delta &lt; 1.-1 &lt; x+2&lt; 1 \mbox{ that implies } -2 &lt; x+1 &lt; 0 \mbox{ therefore } \frac{-1}{2} &gt; \frac{1}{x+1} &gt; 0. This cannot be correct as the answer for delta is \frac{\epsilon}{1+\epsilon}. My algebra is rusty. Please help, thanks.

 

[snipez90]

Not sure why you abandoned your initial line of thought, since there is no single correct answer for delta. The only problem with your analysis is that you have -2 < x + 1 < 0 and then took reciprocals, but you have 0 on one side so following the normal operations of taking reciprocals of inequalities would yield 1/(x+1) > 1/0, which is absurd. Instead, suppose you had picked delta \leq 1/2. Then all that would change is that you now have -1/2 < x + 2 < 1/2 so -3/2 < x + 1 < -1/2, which means that 1/2 < |x+1| < 3/2 and finally 2/3 < 1/|x+1| < 2. Now you have an upper bound on 1/|x+1| without having to worry about division by zero.

 

[John O' Meara]

Using snipez90's suggestion for delta, I get \delta=min(\frac{1}{2},\frac{\epsilon}{2}). I was wondering how the book got the result I said it got, that is why I abandoned my original line of thought.

 

[arildno]

John:

When it comes to estimates and inequalities like this, you are NOT to reproduce the "answer" the book gives you (I find it scandalous and deeply anti-pedagogic that a textbook actually provides "answers" to such exercises!)

Depending on how you think, numerous CORRECT deductions might yield very different delta-values, as functions of the epsilon.

It can be that we may find the "optimal" (i.e, largest) delta that CAN work, but this does not in any way change the validity of a deduction leading to a smaller delta.

Thus, there is no single, correct answer to these types of questions (there are many correct answers)

 

[srijithju]

I am not very familiar with the delta epsilon proof , but is it not sufficient to show that for a given delta you can find an epsilon , do we need to show the opposite ?

 

[John O' Meara]

I am sorry if I broke one of your rules. It was not my intension to do so. As I am teaching myself from a University maths book, which are generally not designed with the self taught in mind. I find that the answer to odd numbered questions is in general the only guide as to how I might be progressing. I admit that that is not ideal and that having some answers to questions can throw you (as it did here) off a valid line of reasoning or investigation. All I can do is offer my apology again.

I am just learning the epsilon delta proof myself. Thanks for the replies.