# Standard limit proof

For any x > 0, we have lim n-->infinity of x^1/n = 1

Hint: Treat cases x >= 1 and x < 1 separately. You might wish to first use the following to prove the preliminary result that for every epsilon > 0, and every real number M > 0, there exists an n such that M^1/n <= 1+epsilson:

"Let x be a real number. Then the limit lim n-->infinity of x^n exists and is equal to zero when the absolute value of x < 1, exists and is equal to 1 when x=1, and diverges when x= -1 of when absolute value of x > 1."

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