The following came from a step in my EM homework. I came up with the answer that all of my resources (calculator, WolframAlpha, and a friend) were indicating as the correct answer, so it is not about the homework. This is about the permissibility of what was done.(adsbygoogle = window.adsbygoogle || []).push({});

1. [tex]x=\lim_{n\rightarrow\infty}n\sin\frac{\pi}{n}[/tex]

2. [tex]x=\lim_{n\rightarrow\infty}\frac{n}{\csc\frac{\pi}{n}}[/tex]

3. [tex]x=\lim_{n\rightarrow\infty}\frac{1}{\frac{\pi\cos\frac{\pi}{n}}{n^2\sin^2\frac{\pi}{n}}}[/tex]

4. [tex]x=\lim_{n\rightarrow\infty}\frac{n^2\sin^2\frac{\pi}{n}}{\pi\cos\frac{\pi}{n}}[/tex]

5. [tex]x=\lim_{n\rightarrow\infty}\frac{x^2}{\pi\cos\frac{\pi}{n}}[/tex]

6. [tex]\frac{1}{x}=\lim_{n\rightarrow\infty}\frac{1}{\pi\cos\frac{\pi}{n}}[/tex]

7. [tex]x=\lim_{n\rightarrow\infty}\pi\cos\frac{\pi}{n}[/tex]

8. [tex]x=\pi\cos0=\pi[/tex]

Given that the exponent on x in step 5 is not 1, x is not zero or infinity, and that x is not the entirety of the argument of the limit, can we/I conclude that this will work in general?

Other possible solutions between 1 and 8 include a change in variable, the limit if the series expansion at infinity, the squeeze theorem, or using the inverse of step two instead.

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# Permissibility of Limit Technique

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