A limit

limnsin(n3n)n lim_{n \rightarrow \infty} \sqrt[n]{sin(\frac{n}{3^n})} , while n is a natural number

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

so the above limit can be yielded by limnsin(n+13n+1)sin(n3n)lim_{n \rightarrow \infty} \frac{sin(\frac{n+1}{3^{n+1}})}{sin(\frac{n}{3^n})}

limnsin(n33n+133n)sin(n3n)lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n} +\frac{1}{3*3^n})}{ sin(\frac{n}{3^n})}

limnsin(n33n)cos(133n)+cos(n33n)sin(133n)sin(n3n)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})}

limnsin(n33n)cos(133n)sin(n3n)+cos(n33n)sin(133n)sin(n3n)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})}

limnsin(n33n)cos(133n)sin(n3n)+0 lim_{n \rightarrow \infty} \frac{sin(\frac{n}{3*3^n})cos(\frac{1}{3*3^n})}{sin(\frac{n}{3^n})}+0

assume n3n=u\frac{n}{3^n} =u ,thus

limu0sin(13u)1sin(u)=13lim_{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
4 years, 10 months ago

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