21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for an...

Question

21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for any positive integer n, even x^(1/1,000,000).

Accepted Answer

Welcome back, everyone. Which function grows faster as X approaches infinity? LN of X or X erases the power of 1 divided by 100,000. A says LN of X and B says X to the power of 1 divided by 100,000. So for this problem, let's remember that if we want to compare the growth of two functions, we want to identify the limit as X approaches infinity of their ratio. What we're going to do is simply use LN of X in the numerator. And we're going to use X to the power of 1 divided by 100,000. In the denominator. So now as x approaches infinity, both the numerator and the denominator are going to approach infinity. And what we can do is simply apply Lapito's rule, right? Because we have an indeterminate form infinity divided by infinity, which is suitable for Lapito's rules to be applied. According to Lapto's rule, we want to find the derivative of the numerator. The derivative of LN of X is one divided by X, and we also want to differentiate our denominator. In the denominator, we're going to apply the power rule. And we are going to get X to the power of 1 divided by 100,000 minus 1, right? And now let's go ahead and simplify. This is equal to limit as x approaches infinity of 1 divided by X, right, which is multiplied by X to the power of 1 divided by 100,000 minus 1. And using the properties of exponents, we can basically conclude that this is limit as X approaches infinity of 1 divided by X to the power of 1. Divided by 100,000. Because we have 1 the power of 1 plus. -1, right? When we multiply X by x the power of 1 divided by 100,000 minus 1, we're adding the exponents. 1 minus 1 gives us 0, and we're left with the original exponent. And we already know that the denominator function approaches infinity when x approaches infinity. So we get 1 divided by infinity. A constant divided by an infinitely large number gives us a limit of 0. Whenever we get a limit of 0, it essentially implies that the denominator function dominates over the numerator function and grows faster. So the correct answer for this problem corresponds to the answer choice B. Thank you for watching.

21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for any positive integer n, even x^(1/1,000,000).

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