Question

pls solve itt as fast​

Answer 1

Answer:

e^(-1/2)

Step-by-step explanation:

$\displaystyle\lim_{n\rightarrow0}(\cos n)^{1/n^2}\\\\=\exp\left(\lim_{n\rightarrow0}\ln\left((\cos n)^{1/n^2}\right)\right)$

Working on the inner part:

$\displaystyle\lim_{n\rightarrow0}\ln\left((\cos n)^{1/n^2}\right)\\\\=\lim_{n\rightarrow0}\frac{\ln(\cos n)}{n^2}$

This tends to the form 0/0, so we use L'Hospital's Rule:

$\displaystyle\lim_{n\rightarrow0}\frac{\ln(\cos n)}{n^2}\\\\=\lim_{n\rightarrow0}\frac{\frac1{\cos n}\times-\sin n}{2n}\\\\=\lim_{n\rightarrow0}\frac{-1}{2\cos n}\times\lim_{n\rightarrow0}\frac{\sin n}{n}\\\\=\tfrac{-1}{2}\times1\\\\=-\tfrac12$

Therefore

$\displaystyle\lim_{n\rightarrow0}(\cos n)^{1/n^2}\ =\ e^{-1/2}$

Answer 2

Answer:

1/√e

Step-by-step explanation:

I did it same way but messier. :D