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 December 4th, 2011, 06:45 PM #1 Member   Joined: Nov 2011 Posts: 35 Thanks: 0 Positive and negative results Hello, I have a limit: $\lim_{x \to 0}\frac{ (1-cos(x))^{2} }{ tan^{3}(x) - sin^{3}(x) }$ I've simplified it to such view: $\lim_{x \to 0} \frac{x^{2}/2 \cdot x^{2}/2}{x^{3} \cdot x^{2}/2}$ Wolframalpha shows, that there are 2 results: ±inf: http://www.wolframalpha.com/input/?i...%2C+x-%3E0+%29 I can't understand, why does the result have ±?
 December 4th, 2011, 06:55 PM #2 Newbie   Joined: Jul 2011 Posts: 5 Thanks: 1 Re: Positive and negative results Wolfram has showed these results as the limit of x->0 has a different result depending which side you approach the limit from.
 December 4th, 2011, 06:57 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Positive and negative results We are given: $\lim_{x\to 0}\frac{$$1-\cos x$$^2}{\tan^3x-\sin^3x}=\lim_{x\to 0}\frac{$$1-\cos x$$^2}{\tan^3x$$1-\cos^3x$$}=\lim_{x\to 0}\frac{$$1-\cos x$$^2}{\tan^3x$$1-\cos x$$$$1+\cos x+\cos^2x$$}=$ $\lim_{x\to 0}\frac{1-\cos x}{\tan^3x$$1+\cos x+\cos^2x$$}=\lim_{x\to 0}\frac{1-\cos x}{$$1-\cos^2x$$\sin x\sec^3x$$1+\cos x+\cos^2x$$}=$ $\lim_{x\to 0}\frac{\cos^3x}{$$1+\cos x$$\sin x$$1+\cos x+\cos^2x$$}$ We have then: $\lim_{x\to 0^{+}}\frac{\cos^3x}{$$1+\cos x$$\sin x$$1+\cos x+\cos^2x$$}=+\infty$ $\lim_{x\to 0^{-}}\frac{\cos^3x}{$$1+\cos x$$\sin x$$1+\cos x+\cos^2x$$}=-\infty$
 December 4th, 2011, 06:57 PM #4 Member   Joined: Nov 2011 Posts: 35 Thanks: 0 Re: Positive and negative results So, can I just calculate it and leave the answer as: inf? Would it be right?
 December 4th, 2011, 07:03 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Positive and negative results I would simply state that the limit does not exist. There is a vertical asymptote of x = 0. On the left the function goes to negative infinity and on the right it goes to positive infinity, much like the function f(x) = 1/x.
 December 4th, 2011, 07:04 PM #6 Member   Joined: Nov 2011 Posts: 35 Thanks: 0 Re: Positive and negative results Okay, thanks for the tip The question is solved.

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