Squeeze theorem - Finding Sequences to Squeeze n n n1/n and...

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Finding Sequences to Squeeze n n + n 1 /n and Hence, to Compute lim n →∞ parenleftbigg n n + n 1 /n parenrightbigg +L’Hˆopital’s Rule Does Work JO We have proven lim n →∞ n 1 /n = 1 . This means that for any ε > 0 there is a positive integer M such that if n M then | n 1 /n - 1 | ≤ ε. I don’t need to worry about the absolute value signs, because n 1 /n - 1 > 0 , so | n 1 /n - 1 | = n 1 /n - 1 & n 1 /n 1 . I can pick any positive value for ε and I am guaranteed an M (that I don’t have to find) such that for n M then n 1 /n - 1 = | n 1 /n - 1 | ≤ ε. 1
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Out of the air, I let ε = 1 / 10, so there is an M 1 such that 1 n 1 /n < 11 / 10 , n M. Note: 11 / 10 = 1 + ε. Now I build the sequence, add n to each term and divide each term by n. 1 + n n n + n 1 /n n n + 11 / 10 n , n M. Next invert each term, recall 0 <a b then 1 /b 1 /a. n n + 11 / 10 n n + n 1 /n n 1 + n , n M. Finally, it is easy to see that lim n →∞ parenleftbigg n n + 11 / 10 parenrightbigg = lim x →∞ parenleftbigg x x + 11 / 10 parenrightbigg = lim x →∞ 1 1 = 1 . It is easier to see lim n →∞ n 1 + n = 1 . Thus the Squeeze Theorem 2 says lim n →∞ n n + n 1 /n = 1 . Another Approach. L’Hˆ opital’s Rule It not so bad. lim n →∞ parenleftbigg n n + n 1 /n parenrightbigg = lim x →∞ parenleftbigg
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