By HarrySummers June 23, 2011
I have used another Technician for years and just got refered to Mobile computer and THEY ROCK! My son appreciate his ps3 being repaired and i appreciate the virus cleaning, once again Thank you ...read more
By Cindy March 03, 2011
This store is horrible on customer service. They do not greet you or acknowledge that you are even in the store. Many times they act like you are interrupting them to wait on you. I have actually had wait while they carrried on personal phone calls and then look at you like you are eavesdropping on their conversation. They are rude and down right disrepectful. I have been in the customer service business for a long time and this is the worst I have ever seen. ...read more
The definition of a limit continues on in our learning limits and how they are applied. Specifically we are going to look at some proofs in this section. With a precise definition we can now prove many limit properties. Definition of a Limit The \(\delta\) value will always depend on the \(\epsilon\) value. That is important to remember in order to understand what is happening in theselimits. Once you find a value that works in your problem then smaller values of that \(\delta\) also work. Problem 1. For the given function f(x) and values L,c, and \(\epsilon\) > 0 find the largest open interval about c on which the inequality |f(x)-L| < \(\epsilon\) holds. Then determine the largest value for \(\delta\) >0 such that 0<|x-c|<\(\delta\)\(\rightarrow\) |f(x)-L|<\(\epsilon\). f(x) = \(\frac{1}{x}\), L= 0.125, c = 8, \(\epsilon\) = 0.015 Solve |f(x) - L| < \(\epsilon\) to find the largest interval containing c on which the inequality holds. Write the inequality without absolute value. Substitute \(\frac{1}{x}\) for f(x), 0.1215 for L, and 0.015 for \(\epsilon\). |f(x)-L| < \(\epsilon\) |\(\frac{1}{x}\) - 0.125| < 0.015 -0.015 < \(\frac{1}{x}\) - 0.125 < 0.015 Simplify by adding 0.125 to all expressions. -0.015 < \(\frac{1}{x}\) - 0.125 < 0.015 0.11 < \(\frac{1}{x}\) < 0.14 Simplify further by taking the reciprocals of all terms. Be sure to reverse the inequalities. Make sure you check that the rounded endpoints still allow the inequality |f(x)-L| < \(\epsilon\) to hold true. 9.0909 > x > 7.1429 Since the interval 7.1429 < x < 9.0909 is not centered on c=8, \(\delta\) is the distance from 8 to closer endpoint of the interval. The value of \(\delta\) is then 0.8571. Problem 2. For the given function \(f(x)\), the point \(c\), and a positive number \(\epsilon\), find \(L=lim f(x)\). Then find a number \(\delta\) > 0 such that for all x, 0<|x-c|<\(\delta\)\(\rightarrow\)|\(f(x)-L\)| < \(\epsilon\). \(f(x)=8-3x, c=4, \epsilon=0.03\) Notice that \(f(x)\) is a linear function. Since a linear function is a simple polynomial function and it is defined for all \(x\), the limit \(L\) = lim is the value of \(f(x)\) at \(c\). Evaluate the function at \(c=4\). \(L=8-3(4)=-4\) To find \(\delta\), begin by solving the inequality \(|f(x)-L|<\epsilon\) to find an open interval \((a,b)\) containing \(c\) on which the inequality holds for all \(x eq c\). Remove the absolute value sign and rewrite the inequality as a compound inequality. \(|(8-3x)-(-4)| < 0.03\) \(-0.03 < (8-3x) - (-4) < 0.03\) Simplify the center expression and isolate the x-term. \(-0.03 < -3x+12 < 0.03\) \(-12.03 < -3x < -11.97\) Then isolate \(x\) in the center. Notice that the direction of the inequalities has been changed. \(4.01 > x > 3.99\) Therefore, for \(x\) in the interval \((3.99,4.01)\), the inequality \((8-3x)-(-4)|<0.03\) holds. Now choose a positive value for \(\delta\) that places the open interval \((c-\delta,c+\delta)\) centered on \(c\) inside the interval \((3.99,4.01)\). The distance to the first endpoint is \(4-3.99=0.01\). The distance to the second endpoint is \(4.01 -4=0.01\). The largest possible value for \(\delta\) is \( 0.01\). Therefore, for all \(x\) satisfying \(0<|x-4|<0.01\), the inequality \(|(8-3x)-(-4)|<0.03\) holds. Problem 3. Give an \(\epsilon - \delta\) proof of the limit fact. \(lim_{x \to 0}(5x+1)=1\) We begin by giving the precise meaning of a limit. To say that \(lim f(x)=L\) means that for each \(\epsilon>0\) there is a corresponding \(\delta>0\) such that \(|f(x)-L|<\epsilon\), provided that \(0<|x-c|<\delta\). To establish the proof, we first perform a preliminary analysis to find the appropriate choice of \(\delta\). Let \(\epsilon\) be any positive number. We must produce a \(\delta > 0\) such that \( 0<|x-0|<\delta \rightarrow |(5x+1) -1 < \epsilon\). By simplifying the inequality on the right, we will determine the value of \(\delta\) needed for the inequality on the left. \(|(5x+1)-1| < \epsilon \Leftrightarrow |5x| < \epsilon \) \(\Leftrightarrow |5| |x| < \epsilon\) \(\Leftrightarrow 5|x| < \epsilon\) \(\Leftrightarrow |x| < \frac{\epsilon}{5}\). Now we see that we should choose \(\delta=\frac{\epsilon}{5}\). We can now construct the formal proof of the statement \(\lim_{x \to 0}(5x+1)=1\). Let \(\epsilon > 0\) be given. Choose \(\delta=\frac{\epsilon}{5}\). Then \(0<|x-0|<\delta\) implies the following chain of equalities and an inequality. \(|(5x+1)-1| = |5x|\) Simplify inside the absolute value bars. \( = |5| |x|\) Rewrite as product of absolute values. \( = 5 |x| \) Simplify. \( =5 |x-0|\) Rewrite expression inside absolute value bars. \( < 5\delta\) Apply the condition \(0<|x-0|<\delta\). \( = \epsilon \) Substitute \(\delta=\frac{\epsilon}{5}\). The result of the above chain of equalities and an inequality is that \(|(5x+1)-1|< \epsilon\). Therefore, we have proven that \(\lim_{x \to 0}(5x+1)=1\). Conclusion The definition of a limit has some nice proofs that can establish. If you go over these problems slowly the ideas behind limits will begin to emerge. While this section is short it is essential to understanding limits and beginningcalculusin general. ...read more
By Jason's Computer Repair, Video, and Audio Services February 26, 2018
This is my Step by Step Excel 2016 Tutorial. Excel 2016 is new and it can be confusing. I was told this recently and I totally agree with that statement. It looks and feels different than the previous versions. If you have always been a little afraid or unsure of how to use Microsoft Excel well, I wrote this step by step Excel 2016 tutorial just for you! After reading this and going through my examples I promise you will have a much better grasp on how to use Excel spreadsheets. Read the rest at :http://aindien.com/step-by-step-excel-2016-tutorial/ ...read more
By Jason's Computer Repair, Video, and Audio Services August 26, 2017
Network Basics: Easy Addressing and Routing is all about the TCP/IP protocols and what they do. Our present day internet is totally dependent on this system because because of how it developed and without a doubt it can be confusing. Think about it: The reason TCP/IP was used was because it was rock solid. However, it seems like there are a million details to it and I am sure you would agree with me. The addressing and routing that moves all of our internet traffic between Facebook, Twitter, and all of our other favorite websites uses TCP/IP primarily so there has to be a reason for it right? There really is so let me explain to it to you. Easy Addressing and Routing IP addressing is the system that dictates how every device communicates with each other over a private network or the internet as a whole. Every device must have a unique name and be identified thusly so all our computers who to talk to. If you find this article interesting then check out the full page at: http://aindien.com/network-basics-easy-addressing-and-routing/ ...read more
By Jason's Computer Repair, Video, and Audio Services August 21, 2017
Many people choose to watch their favorite television shows and movies by streaming them over the Internet, but would prefer a larger screen or more comfort while watching. For this reason, some people have connected their PC to the television, allowing them to watch streaming video from the comfort o... ...read more
There are federal rules and regulations that must be followed pertaining to the use of electronics on airplanes. Certain electronics give up signals that interfere with the flight navigation system used to monitor the activity of airplanes. For this reason, it is best to know the rules before flying. ...read more
