pisrationalhahaha
New member
- Joined
- Aug 22, 2017
- Messages
- 46
Using limit definition for 1/n×cos(npi/4)
\(\displaystyle \lim_{n\rightarrow +\infty }\frac{1}{n}cos\frac{n\pi }{4}=0\)
I want to prove this limit using the limit definition
What I did is:
\(\displaystyle \left | \frac{1}{n}cos\frac{n\pi }{4}-0 \right |< \varepsilon\)
Then \(\displaystyle \left | \cos\frac{n\pi }{4}-0 \right |< n\varepsilon\)
But
\(\displaystyle 0< \left | \cos\frac{n\pi }{4} \right |< 1\)
Then it is sufficiant to say that \(\displaystyle 1< n\varepsilon\)
So that \(\displaystyle n> \frac{1}{\varepsilon }\)
Let \(\displaystyle n_0 =\left [ \frac{1}{\varepsilon } \right ]+1\)
(integer part)
Therefore \(\displaystyle \forall \varepsilon > 0,\exists n_0\in \mathbb{N}|\forall n> n_0,\left |\frac{1}{n}cos\frac{n\pi }{4}-0 \right |< \varepsilon\)
Is what I did correct ?
And for mathematical language, is my syle of writing correct ?
\(\displaystyle \lim_{n\rightarrow +\infty }\frac{1}{n}cos\frac{n\pi }{4}=0\)
I want to prove this limit using the limit definition
What I did is:
\(\displaystyle \left | \frac{1}{n}cos\frac{n\pi }{4}-0 \right |< \varepsilon\)
Then \(\displaystyle \left | \cos\frac{n\pi }{4}-0 \right |< n\varepsilon\)
But
\(\displaystyle 0< \left | \cos\frac{n\pi }{4} \right |< 1\)
Then it is sufficiant to say that \(\displaystyle 1< n\varepsilon\)
So that \(\displaystyle n> \frac{1}{\varepsilon }\)
Let \(\displaystyle n_0 =\left [ \frac{1}{\varepsilon } \right ]+1\)
(integer part)
Therefore \(\displaystyle \forall \varepsilon > 0,\exists n_0\in \mathbb{N}|\forall n> n_0,\left |\frac{1}{n}cos\frac{n\pi }{4}-0 \right |< \varepsilon\)
Is what I did correct ?
And for mathematical language, is my syle of writing correct ?
