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 October 26th, 2019, 11:29 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math A limit Evaluate $\displaystyle \lim_{x\rightarrow 0 } \frac{x^{x+1} -\sin(x)}{x-\tan(x)}$. October 26th, 2019, 05:24 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 Does not converge. (Goes towards $+\infty$ from the right and $-\infty$ from the left.) Thanks from topsquark October 26th, 2019, 10:02 PM   #3
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 Originally Posted by DarnItJimImAnEngineer Does not converge. (Goes towards $+\infty$ from the right and $-\infty$ from the left.)
I'm going to poke my nose in here. How do you handle the limit for negative x as $\displaystyle x^{x + 1}$ is a bit psychotic for negative x?

-Dan October 27th, 2019, 12:37 AM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math $\displaystyle x\rightarrow 0$. $\displaystyle l=\lim_{x\rightarrow 0 } \frac{x^{x+1} -\sin(x)}{x-\tan(x)}=\lim x^{x+1}(x-\tan(x))^{-1} - \lim \sin(x)/[x-\tan(x)]=\lim x^{x+1}/[x-\tan(x)]-3$. $\displaystyle l=\lim -\frac{1}{\sin(x)}-3=-\infty$. Last edited by skipjack; November 2nd, 2019 at 11:34 PM. October 27th, 2019, 08:21 AM   #5
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 Originally Posted by topsquark I'm going to poke my nose in here. How do you handle the limit for negative x?
Very lazily. I let MATLAB handle the calculations, and the real portion was antisymmetric (i.e., odd). On inspection, the imaginary part, as chosen by MATLAB, goes infinite positive as $x\rightarrow 0^-$. October 27th, 2019, 08:36 AM   #6
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 Originally Posted by idontknow $\displaystyle \lim \sin(x)/[x-\tan(x)] = 3$.
Sorry, but I don't know how you got that. $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x - \tan(x)}$ is unbounded.

-Dan

Last edited by skipjack; November 2nd, 2019 at 11:36 PM. October 28th, 2019, 10:31 PM   #7
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 Originally Posted by topsquark Sorry, but I don't know how you got that. $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x - \tan(x)}$ is unbounded. -Dan
My mistake:

$\displaystyle \lim_{x\rightarrow 0} \frac{d\sin(x)}{dx-d\tan(x)}=\lim_{x\rightarrow 0} \frac{1}{1-1/[ \cos^{2}(x)]}=\infty$.

Last edited by skipjack; November 2nd, 2019 at 11:37 PM. November 2nd, 2019, 04:50 AM   #8
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 Originally Posted by idontknow My mistake: $\displaystyle \lim_{x\rightarrow 0} \frac{d\sin(x)}{dx-d\tan(x)}=\lim_{x\rightarrow 0} \frac{1}{1-1/[ \cos^{2}(x)]}=\infty$.
Careful here. Derivative of sin(x) is not 1.

Last edited by skipjack; November 2nd, 2019 at 11:38 PM. November 2nd, 2019, 05:10 AM   #9
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 Originally Posted by RDKGames Careful here. Derivative of sin(x) is not 1.
Not again! My mistake...

Last edited by skipjack; November 2nd, 2019 at 11:38 PM. November 2nd, 2019, 02:35 PM   #10
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Quote:
 Originally Posted by RDKGames Careful here. Derivative of sin(x) is not 1.
Oops! I missed that one myself!

Good catch.

-Dan

Last edited by skipjack; November 2nd, 2019 at 11:39 PM. Tags limit 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 zylo Calculus 13 May 31st, 2017 01:53 PM veronicak5678 Real Analysis 4 August 22nd, 2011 11:07 AM panky Calculus 1 August 8th, 2011 04:14 PM panky Calculus 9 July 22nd, 2011 05:11 PM conjecture Calculus 1 July 24th, 2008 02:14 PM

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