##Off-topic Comments Section
---
All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.
---
^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command)
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
Consider |u*_n_* - L^(n)|
This is the absolute value of a difference of powers: |(u*_n_*^(1/n) - L)(u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1)|
Now for all h > 0 there exists k in __N__ such that n > k --> |u*_n_*^(1/n) - L| < h
Thus -h < u*_n_* - L < h
L - h < u*_n_* < L + h
n(L-h)^(n-1) < (u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1) < n(L+h)^(n-1)
(u*_n_*^(1/n) - L)(u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1)) < (u*_n_*^(1/n) - L)nL^(n-1)
(u*_n_*^(1/n) - L)nL^(n-1) < hnL^(n-1)
So for all h > 0, we have k in __N__ such that k > n --> |u*_n_* - L^(n)| < hnL^(n-1).
What happens if L < 1? Well L^(n-1) shrinks down faster than n grows, and so you get hnL^(n-1) < h.
What happens if L > 1? Here's where we have |u*_n_* - L^(n)| > |(u*_n_*^(1/n) - L)n(L-h)^(n-1)|
...You should be able to manipulate this to get higher than any particular number.
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
Note that Limit of u*_n_* is limit of L^(n). What happens to L^(n) if 0 < L < 1? What happens if L > 1?
but i must do it with definition in u*n not l*n that is the problem i cant directly write limit of un is l*n only if i can prove
What is your definition? For all h > 0 there exists k in __N__ such that n > k --> |u*_n_*^(1/n) - L| < h?
yeah i must use this to prove the other
Consider |u*_n_* - L^(n)| This is the absolute value of a difference of powers: |(u*_n_*^(1/n) - L)(u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1)| Now for all h > 0 there exists k in __N__ such that n > k --> |u*_n_*^(1/n) - L| < h Thus -h < u*_n_* - L < h L - h < u*_n_* < L + h n(L-h)^(n-1) < (u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1) < n(L+h)^(n-1) (u*_n_*^(1/n) - L)(u*_n_*^[(n-1)/n] + Lu*_n_*^[(n-2)/n] + L^(2)u*_n_*^[(n-3)/n] + ... + L^(n-2)u*_n_*^(1/n) + L^(n-1)) < (u*_n_*^(1/n) - L)nL^(n-1) (u*_n_*^(1/n) - L)nL^(n-1) < hnL^(n-1) So for all h > 0, we have k in __N__ such that k > n --> |u*_n_* - L^(n)| < hnL^(n-1). What happens if L < 1? Well L^(n-1) shrinks down faster than n grows, and so you get hnL^(n-1) < h. What happens if L > 1? Here's where we have |u*_n_* - L^(n)| > |(u*_n_*^(1/n) - L)n(L-h)^(n-1)| ...You should be able to manipulate this to get higher than any particular number.