 My Math Forum Limit of f(x)/q(x) with q(c) = 0 that can not be simplified.
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 July 4th, 2019, 05:09 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 Limit of f(x)/q(x) with q(c) = 0 that can not be simplified. Is it true that if $\displaystyle \lim_{x \to c} \frac{f(x)}{q(x)}$ with q(c) = 0, $\displaystyle f(c) \in R$ and if that fraction can not be simplified any more, then this limit will always diverge. Example: $\displaystyle \lim_{x \to 1} \frac{1}{x-1}$ diverges because: $\displaystyle \lim_{x \to 1^-} \frac{1}{x-1} = - \infty$ $\displaystyle \lim_{x \to 1^+} \frac{1}{x-1} = \infty$ July 4th, 2019, 08:52 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,554 Thanks: 1403 no $\lim \limits_{x \to 0} \dfrac{\sin(x)}{x} = 1$ Thanks from topsquark and babaliaris July 4th, 2019, 09:11 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra On the other hand, if $f(x)$ has a Taylor series $T(f, c)$ about $x=c$ and $q(x)$ has a Taylor series $T(q, c)$ about $x=c$ and $\frac{T(f,c)}{T(q,c)}$ cannot be simplified, I think that the limit won't exist, but you won't always have different one-sided limits. (e.g. $\frac1{x^2}$). Thanks from topsquark July 4th, 2019, 11:46 AM   #4
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,685
Thanks: 2666

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by babaliaris Example: $\displaystyle \lim_{x \to 1} \frac{1}{x-1}$ diverges because: $\displaystyle \lim_{x \to 1^-} \frac{1}{x-1} = - \infty$ $\displaystyle \lim_{x \to 1^+} \frac{1}{x-1} = \infty$
$\frac1{x-1}$ doesn't converge to a limit as $x\to1$ but not because the one-sided limits differ. It's because they don't exist.

$\displaystyle \lim_{x \to 1^-} \frac{1}{x-1} = -\infty$ doesn't mean that there is a limit with a value of $-\infty$. It's a shorthand for saying that the function grows without bound in the negative direction - which means that the limit doesn't exist. July 4th, 2019, 09:31 PM   #5
Global Moderator

Joined: Dec 2006

Posts: 20,978
Thanks: 2229

Quote:
 Originally Posted by babaliaris . . . can not be simplified any more
How is that property defined? July 5th, 2019, 12:04 PM   #6
Senior Member

Joined: Oct 2015
From: Greece

Posts: 137
Thanks: 8

Quote:
 Originally Posted by skipjack How is that property defined?
I wasn't trying to specify a rule, just an observation. "Can not be simplified any more", it wasn't my intention to specify this as a property.

Last edited by skipjack; July 7th, 2019 at 12:59 AM. July 6th, 2019, 01:45 PM   #7
Senior Member

Joined: Sep 2016
From: USA

Posts: 647
Thanks: 412

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by babaliaris I wasn't trying to specify a rule, just an observation. "Can not be simplified any more", it wasn't my intention to specify this as a property.
Skipjack's point is that the phrase "can not be simplified any more" does not have any meaning. So nobody knows how to answer your question. This would be like me asking you to solve $3y + 4x = 7$ when $x$ is green. It is meaningless.

Last edited by skipjack; July 7th, 2019 at 12:59 AM. Tags fx or qx, limit, simplified Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post life24 Pre-Calculus 3 January 15th, 2017 11:30 AM iScience Algebra 1 August 31st, 2016 12:56 PM Dalekcaan1963 Algebra 2 December 12th, 2015 08:57 PM mms Algebra 7 November 2nd, 2011 08:08 AM empiricus Algebra 13 April 27th, 2010 06:55 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      