/sci/ - Science & Math » Thread #11076633

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Anonymous Sun 20 Oct 14:04:33 2019 No.11076633 View Reply Original Report

Quoted By: >>11076687 >>11077934 >>11078310 >>11078322 >>11078405

can someone please, for the love of god help me with this problem

>real analysis class

>getting fucked rekt right now

Anonymous

Anonymous Sun 20 Oct 2019 14:10:21 No.11076649 Report

Quoted By: >>11076657

Doesn't exist. pi/x has no limit as x goes to 0: it approaches +inf from above and -inf from below.

Anonymous

Anonymous Sun 20 Oct 2019 14:13:14 No.11076657 Report

Quoted By: >>11076661 >>11076677

>>11076649

i know it doesnt exist, but somehow i am suppose to prove it by finding some Xn taking that limit as x -> goes to infitiny, then subisiting that in the x for the function cos(pi/x)

Anonymous

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Anonymous Sun 20 Oct 2019 14:14:19 No.11076661 Report

Quoted By: >>11078322 >>11079557

>>11076657

Proof should look like this

Anonymous

Anonymous Sun 20 Oct 2019 14:20:14 No.11076677 Report

Quoted By: >>11078329

>>11076657
take xn = 1/n.

cos(1/x(2n)) = cos (2n pi) = 1
cos(1/x(2n+1)) = cos (2n pi + pi) = - 1

As xn is a sequence that converges to 0 and we found two subsequences for which the limits are different, there is no limit to your function

Anonymous

Anonymous Sun 20 Oct 2019 14:22:29 No.11076687 Report

Quoted By: >>11077968

>>11076633
Lol this is such a simple problem. In my first year undergrad I wouldn't even get 4 points for this, it would be just a minor part in a larger problem. Consequences of studying at a shit school, I guess.

Anonymous

Anonymous Sun 20 Oct 2019 22:27:28 No.11077934 Report

Quoted By: >>11077963

>>11076633
This limit does not exist

Anonymous

Anonymous Sun 20 Oct 2019 22:38:51 No.11077963 Report

Quoted By:

>>11077934
BASED

Anonymous

Anonymous Sun 20 Oct 2019 22:42:18 No.11077968 Report

Quoted By: >>11079786

>>11076687
cool but do you have sex regularly

Anonymous

Anonymous Mon 21 Oct 2019 00:42:12 No.11078310 Report

Quoted By: >>11078405

>>11076633
Limit doesn't exist, it just keeps oscillating between -1 and 1. Kind of annoying to prove. Just assume that a limit exists and then proof by contradiction the epilson-delta definition.

Anonymous

Anonymous Mon 21 Oct 2019 00:45:03 No.11078322 Report

Quoted By:

>>11076633
You shouldn’t have posted this
>>11076661
faggot.

Anonymous

Anonymous Mon 21 Oct 2019 00:46:50 No.11078329 Report

Quoted By:

>>11076677
this or Taylor/Laurent series is the most simple method

Anonymous

Anonymous Mon 21 Oct 2019 01:12:53 No.11078405 Report

Quoted By: >>11079395

>>11076633
>>11078310
You need the epsilon delta definition. You'll have to show that in an interval around 0, by the archimedean property, there will always exist an interval that beats epsilon which contradicts the statement that that is the closure of the function.

Or at least something like that, I'm learning real analysis too, so if someone wants to correct me, go ahead.

Anonymous

Anonymous Mon 21 Oct 2019 09:44:13 No.11079395 Report

Quoted By:

>>11078405
just use the fact that two subsequences converge to two different limits. It's the lowest IQ way to go about this problem.

Anonymous

Anonymous Mon 21 Oct 2019 11:58:59 No.11079557 Report

Quoted By:

>>11076661
We have that $\lim_{x \to a } f(x) = L$ iff for every sequence $( x_n )$ such that $x_n \to a$ with $x_n \neq x$ for all n, we have $f( x_n ) \to f(x)$ . Let $x_n = \frac{1}{n}$ . Then $f( x_n ) = cos ( n \pi )$ . If the limit exists, then this sequence converges somewhere. From here on, it's easy for you to find 2 subsequences that converge to different limits. Hint: find n's such that $cos( n \pi )$ is constant.

Anonymous

Anonymous Mon 21 Oct 2019 14:25:12 No.11079786 Report

Quoted By:

>>11077968
Typical cumbrain response. Enjoy leaving no legacy LOL

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