{"id":1115,"date":"2017-10-17T12:49:30","date_gmt":"2017-10-17T16:49:30","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=1115"},"modified":"2017-10-17T12:49:30","modified_gmt":"2017-10-17T16:49:30","slug":"definitions-101217","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/10\/17\/definitions-101217\/","title":{"rendered":"Definitions 10\/12\/17"},"content":{"rendered":"

In class we discussed the Sequential Criterion for Functional Limits, Divergence Criterion for functional limits, and Characterizations of continuity. We emphasized that we have built an understanding of functional limits and convergence that allow us to use several definitions interchangeably depending on the proof we are trying to complete.<\/p>\n

Theorem 4.2.3 (Sequential Criterion for Functional Limits)<\/strong><\/p>\n

Given a function \"f\rightarrow and a limit point \"c\" of \"A\", lim\"f(x)=L\" iff for all sequences \"(x_n)\subseteq satisfying \"x_n\neq and \"(x_n)\rightarrow, it follows that \"f(x_n)\rightarrow.<\/p>\n

Pf template:<\/p>\n

Let \"\epsilon, and choose $latex\\delat >0$ so that \"x\in.<\/p>\n

\"\Rightarrow\" Start by assuming\u00a0lim\"f(x)=L\" and consider the arbitrary sequence \"(x_n)\subset such that \"x_n and \"(x_n)\rightarrow. Hence, we know that there exists \"V_\delta with with the property that all\u00a0\"x\in satisfy \"f(x)\in. Notice that \"x_n\" is eventually in \"(V_\delta and thus \"f(x_n)\" is eventually in \"V_\epsilon.<\/p>\n

Notice that this part of the proof uses topological definitions. The next part of the proof is a proof by contradiction.<\/p>\n

\"\Leftarrow\" Now assume that\u00a0lim\"_{x\rightarrow so that there exists an \"\epsilon_\circ such that for all \"\delta>0\" there exists an\u00a0\"x\in such that \"f(x)\notin. We choose \"\delta=\frac{1}{n}\" so that\u00a0\"x\in but\u00a0\"f(x_n)\notin. Hence\u00a0\"f(x_n)\" does not converge to \"L\". Therefore when\u00a0all sequences \"(x_n)\subseteq satisfying \"x_n\neq and \"(x_n)\rightarrow, it follows that \"f(x_n)\" do not converge to \"L\".<\/p>\n

The Divergence Criterion for Functional Limits<\/strong> follows from this: If we can produce two sequences \"(x_n)\" and \"(y_n)\" in \"A\" with \"x_n\neq and \"y_n\neq and lim\"x_n=\"lim\"y_n=c\", but lim\"f(x_n)\neq\"lim\"f(y_n)\", then the functional limit lim\"_{x\rightarrow does not exist.<\/p>\n

We can conclude the results in the Algebraic Limit theorem for Functional Limits by using the definition of functional limits and the Algebraic Limit Theorem for Sequences.<\/p>\n

The Algebraic Limit Theorem:<\/strong><\/p>\n

Let \"f\" and\u00a0\"g\" be functions defined on a domain \"A\subseteq and assume lim\"_{x\rightarrow and\u00a0lim\"_{x\rightarrow. Then<\/p>\n