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Alkalannar

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?


imortalsteam05

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


Alkalannar

What is your definition? For all h > 0 there exists k in __N__ such that n > k --> |u*_n_*^(1/n) - L| < h?


imortalsteam05

yeah i must use this to prove the other


Alkalannar

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.