19. Show that e^x grows faster as x→∞ than x^n for any positiv...

Question

19. Show that e^x grows faster as x→∞ than x^n for any positive integer n, even x^1,000,000. (Hint: What is the nth derivative of x^n?)

Accepted Answer

Welcome back, everyone. Determine whether the following statement is true or false. As x approaches infinity, F of X equals X to the power of 10,000 grows faster than G of X equals e to the power of X. A is as true and B's as false. For this problem, let's begin with an assumption so that we have a reference point. We're going to assume that this statement is true. If this statement is true, then we're going to take the limit as X approaches infinity. We're going to put the function that grows faster, which is F of X in the denominator, and the function that grows slower in the numerator G of X. And make our limit equal to 0, because our denominator dominates over the numerator giving a limit of 0, right? We want to check if this limit is indeed zero. So, what we're going to do is take a limit as X approaches infinity of, it's the power of X divided by X to the power of 10,000. If we simply attempt the right substitution, X approaches infinity, E to the power of infinity is going to approach infinity, and infinity to the power of 10,000 is going to approach infinity. This is an indeterminate form. We cannot sell much. So what we can do is simply apply Lopical's rule. According to Lappito's rule, if we get an indeterminative form of 0 divided by 0 or infinity divided by infinity, we can differentiate. The numerator and we can differentiate the denominator, right? So let's find each derivative. We're going to get limit. As X approaches infinity of E to the power of X, as the derivative in the numerator, and in the denominator we can apply the power rule to get 10,000. X to the power of 9,999. Now, again, we will get the same infinity divided by infinity. We simply reduced our exponent in the denominator, but the numerator will always remain it the power of X. So what we want to do is start with the original function and differentiate. The numerator and the denominator. 10,000 times. So what we're going to get is a limit as X approaches infinity of E to the power of X in the numerator, right? Because no matter how many times we differentiate E to the power of X, we're going to get E to the power of X. But what about the power rule? Well, essentially, if we apply the power rule 10,000 times, we're going to get 10. 1000 pictorial, right? So now notice that we have a constant. In the denominator, so we're now dividing an infinitely large term by a constant which gives us infinity, and infinity is not equal to zero. In fact, if we get infinity, it basically means that the numerator function actually grows faster. So e to the power of X grows faster, meaning the answer to this problem is B false. Thank you for watching.

19. Show that e^x grows faster as x→∞ than x^n for any positive integer n, even x^1,000,000. (Hint: What is the nth derivative of x^n?)

Watch next