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A tricky limit |
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| Aug24-07, 09:31 AM | #1 |
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A tricky limit
1. The problem statement, all variables and given/known data
Find the limit: [tex]lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}}[/tex] 2. Relevant equations 3. The attempt at a solution I first tried to find the limit of ln of the function inorder to turn the power into a multiplication and got: [tex]lim_{x \rightarrow 1} \frac{ ln(2-x) sin \frac{\pi x}{2}}{cos \frac{\pi x}{2}}[/tex] Then I used L'hopitals rule and got: [tex]lim_{x \rightarrow 1} ln( (2-x)^{tan \frac{\pi x}{2}} ) = \pi / 2[/tex] That means that [tex]lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}} = e^{\pi / 2}[/tex] Is that right? I tried putting in values to my calc and it looks like the answer should be 1? What did I do wrong? Thanks. |
| Aug24-07, 10:04 AM | #2 |
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Check your l'Hopital's rule again. I get 2/pi, not pi/2.
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| Aug24-07, 08:59 PM | #3 |
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I also get [itex]2/\pi[/itex], and when I put in x=0.9999 on my calculator, the limit and [itex]e^{2/\pi}[/itex] agree to reasonable accuracy, around 1.89, not 1.
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