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In general, I do not understand when to restrict the value of $\delta $ in limit proofs to less than one or any other real number. The limit definition for a function of two variables in my textbook is given as:

Let f be a function of two variables that is defined on some open disk $B((x_0, y_0);r)$ (if A is in $\mathbb R^n $ and r is a positive number, then the open disk $B(A; r)$ is the set of all points $P$ in $\mathbb R^n$ such that distance between P and A is less than r), except possible at the point $(x_o, y_o)$ itself. Then $$\lim_{(x, y)\to (x_0, y_0)} f(x, y)= L$$ if for any $\epsilon > 0$ there exists a $\delta > 0$ such that if $0< \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta$ then $\left\lvert(f(x, y) - L) \right\rvert < \epsilon$

My question is what is the justification to be able to restrict the value of $\delta$. I tried doing the following example (link at https://drive.google.com/file/d/0B0jd-BnW8J-WOVM2MVd5c3l4TFdMZUFVMlFvRkM0aGxPX3Jz/view?pref=2&pli=1). In this example, I understand that we need to find something to prove $\left\lvert x-3 \right\rvert$ less than the distance and less than $\delta$, so that this can play a part in finding $\epsilon$. What I don't understand is why I can't do it for the other expressions. Also is my proof on the correct path and what steps would I need to take place next?

user278039
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1 Answers1

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Given some $\epsilon>0$, you want to show that you can find/choose a $\delta>0$ (i.e., there exists some $\delta>0$) to satisfy the definition. You have complete freedom to choose this $\delta$ as long as it is positive; declaring it to be $\le 1$ is not a problem.

angryavian
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