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A limit

\( lim_{n \rightarrow \infty} \sqrt[n]{sin(\frac{n}{3^n})} \) , while n is a natural number

theory : if \(X_n \) was a positive Numerical consecutive then the limit \( lim_{n \rightarrow \infty} \sqrt[n]{X_n} =\frac{X_{n+1}}{X_n} \)

so the above limit can be yielded by \(lim_{n \rightarrow \infty} \frac{sin(\frac{n+1}{3^{n+1}})}{sin(\frac{n}{3^n})}\)

\(lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n} +\frac{1}{3*3^n})}{ sin(\frac{n}{3^n})}\)

\(lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n})cos(\frac{1}{3*3^n}) +cos(\frac{n}{3*3^n})sin(\frac{1}{3*3^n}) }{ sin(\frac{n}{3^n})}\)

\(lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n})cos(\frac{1}{3*3^n})}{sin(\frac{n}{3^n})}+\frac{cos(\frac{n}{3*3^n})sin(\frac{1}{3*3^n}) }{ sin(\frac{n}{3^n})}\)

\( lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n})cos(\frac{1}{3*3^n})}{sin(\frac{n}{3^n})}+0\)

assume \(\frac{n}{3^n} =u\) ,thus

\(lim_{u \rightarrow 0}\frac{sin(\frac{1}{3}u)*1}{sin(u)}=\frac{1}{3}\)

this note is for a question by almikdad ismailto show off at his mates lol.

Note by Jafar Badour
1 year, 11 months ago

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