9. True, or false? As x→∞,
g. ln(x) = o(ln(2x))

Question

9. True, or false? As x→∞,

g. ln(x) = o(ln(2x))

Accepted Answer

Welcome back, everyone. True or false. As x approaches infinity, LN of X is equal to a little o of L N X plus 1. A says true, and B says false. For this problem, let's begin by understanding the little O notation. If we have an expression F of X equals O of G of X, it basically means that as X approaches infinity. The limit. As x approaches infinity of F of X divided by G of X is equal to 0, which means that G of X grows faster than F of X. So let's go ahead and calculate the limit as x approaches infinity of F of X, which in this case is LN of X, right, divided by G of X, which is LNX plus 1. If we simply substitute infinity for every X, we're going to get an indeterminate form infinity divided by infinity. So instead we can notice that we have Llan of X in the numerator and the denominator, meaning we can divide all of our terms by LN of X, which leaves us with 1 in the numerator and in the denominator we're going to get 1 plus 1 divided by LN of X. So all that we have to do is understand that as X approaches infinity, LN of X is going to approach infinity as well, right? And then 1 divided by an infinitely large number is going to approach 0 overall. And now using this analysis, we can say that this limit is equal to 1 divided by 1 + 0, which is equal to 1. Because it is not equal to zero, we can conclude that the statement is false and the answer corresponds to the answer choice b. Thank you for watching.

9. True, or false? As x→∞,g. ln(x) = o(ln(2x))

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9. True, or false? As x→∞,
g. ln(x) = o(ln(2x))