
Re: Sequence limit
Posted:
Oct 7, 2013 10:34 AM


On Friday, October 4, 2013 7:36:47 PM UTC5, Bart Goddard wrote: > Mohan Pawar wrote in > > news:@googlegroups.com: > > > > > > >> The only thing that disturbs me about this is that > > >> > you're an "Instructor." > > > > > > I am sorry if my being instructor disturbs you. > > > > "Disturbs?" My classes are full of students who've > > been misled by incompetent "instructors" who say things > > like "Anything to the zero power is 1", which those > > kids bring to university and use to make my job 10 > > times harder. > > > > >However, if error/s in > > > _my solution_ disturb you, show me at which one of the 4 steps you > > > found an error and I will help you. > > > > You're going to help me? Calculus is a freshman class > > and one topic (arguably the MAIN topic) of calculus is > > dealing with indeterminate forms. Not only do you not > > know how to deal with them, you didn't even recognize > > one at hand. Further, you going to let the exponent of the > > expression go to zero, while somehow holding the rest > > of the expression back?
See below under "Somehow explained to Bart Goddard". > > > > Worse, you don't know enough trig to realize that > > sin(x) is zero infinitely often, which means that > > sin(x)^(1/x) is zero infinitely often. So not only > > do you not know calculus, you don't even know trig. > > If you have a high school diploma, you shouldn't. > > And most certainly you shouldn't be an "instructor" > > of math/physics when you don't know any math or > > physics. > > > > > If you plot a simple graph of y= sin x^(1/x), you won't be able to > > > say that "the value of sin x^(1/x) is zero infinitely often". > > > > See those little spikes in your graph? Zoom in on them. > > The more you zoom in, the closer they are to zero. Maybe > > instead of a "simple" graph, you could have tried > > using calculus techniques to create the graph and > > find the coordinates of the cusps...? > > > > > > >> So the that limit can't be 1. > > > > > > This is your conclusion based on wrong assumption that "the value of > > > sin x^(1/x) is zero infinitely often". I don't see any logic in it. > > > > This is my correct concuclusion based on logic. It was the > > logic I gave in the previous post. And above. > > > > So why is it that the most condescending people are > > always the most clueless, oblivious, least trained > > and just plain stupid? Why do they > > usually make those people into deans?
I won't attempt to attain a low level to match you in ranting and displaying vitriol. At times it is better to keep shut and let others wonder that you are stupid than to open your mouth and remove their all doubts. Bart Goddard, when others critically read the solution under "somehow explained to Bart Goddard" you will have removed their doubts about you.
********************************************************* "Somehow" explained to Bart Goddard ********************************************************* I am assuming that the most relevant issue was why limit was decided by the index m and not the base sin(1/m) in lim m > 0 sin (1/m) ^(m).
The original problem was transformed into equivalent problem as below: Find lim m > 0 sin (1/m) ^(m)
Let y = sin (1/m) ^(m) Taking log on both sides => ln(y) = m ln(sin (1/m)) Take limit on both sides as m>0 and evaluating it =>lim m>0 ln(y) = lim m>0 m ln(sin (1/m))= 0 (at the time of evaluating limit, m=0 is the multiplier and one doesn?t need to care about value of ln(sin (1/m)). Also, the limit is determinate.)
=>lim m>0 ln(y) = 0 => lim m>0 y = e^0=1 => lim m>0 sin (1/m) ^(m)= 1 as before. (ALSO, VERIFIABLE ON WOLFRAM ALPHA)
Notice that the new additional steps are no different from my original two line justification that saves above labor except now the index m is brought down as multiplier through log operation. It is still the exponent m now as multiplier that _alone_ decided value of limit. For reference, the original two line justification from my previous post is quoted below:
"Note that the value of sin (1/m) varies from 0 to to 1 BUT exponent m is guaranteed to be zero as m>0. Now if m is replaced by natural number n, the situation does not change sin (1/n)will still be within 0 to 1 and limit will evaluate to due to zero in exponent."
Best regards.
Mohan Pawar Online Instructor, Maths/Physics MP Classes LLC  US Central Time: 9:34 AM 10/7/2013

