# Use limit definition to prove

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

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.

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
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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)

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 ?

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.