# Thread: How do you justify this limit?

1. ## How do you justify this limit?

let {xn} as n tends to infinity be defined by xn=cos(n)/n. The limit is 0, but is the only way to justify this to use the definition of a limit?

2. $\displaystyle -1 \leq cos(n) \leq 1$

$\displaystyle \frac{-1}{|n|} \leq \frac{cos(n)}{n} \leq \frac{1}{|n|}$

as n tends to infinity, you can take n > 0

$\displaystyle lim_{n\to\infty}\frac{-1}{n} = 0$ and $\displaystyle lim_{n\to\infty}\frac{1}{n} = 0$

By the squeeze thereom $\displaystyle \lim_{n\to\infty}\frac{cos(n)}{n} = 0$

3. Originally Posted by amm345
let {xn} as n tends to infinity be defined by xn=cos(n)/n. The limit is 0, but is the only way to justify this to use the definition of a limit?
Usually the definition of the limit is the ultimate authority on a limit's value. How about using the squeeze theorem.

$\displaystyle \frac{-1}{n}\le x_n\le\frac{1}{n}$ for all permissable values. What can you say now?

4. Thank you both! Would I then be able to something similar to show that the limit of (1+1/n)^(n+1) is e?

5. Originally Posted by amm345
Thank you both! Would I then be able to something similar to show that the limit of (1+1/n)^(n+1) is e?
Well, that is a bit of a more tricky situation. You can bound it above and below, but this isn't exactly helpful. Do you know that $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$. If so, just split your limit into $\displaystyle \left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{n}\ri ght)$....and then.............what?

6. ahh I see. perfect! thank you again!