Question

In class, we sketched out the fact that polynomial functions grow slower than exponential functions, which includes statements like n¹⁰⁰ ∈O(eⁿ).

Our method was using l'Hopital's rule "several times," but mathematically speaking, the "several times" part is informal and should instead be written in terms of a proof by induction.
Give a formal proof, using induction, that lim x→[infinity] xᶦ/eˣ =0 for all integers i≥0. (Hint: the base case is i=0 instead of 1 , and for the inductive step, you should be applying l'Hopital's rule exactly one time.)

Answer 1

To prove lim x→[infinity] xᶦ/eˣ = 0 for all integers i≥0 using induction and l'Hopital's rule, we start with the base case i = 0 and show that it holds. Then, assuming it holds for i = k, we prove it for i = k+1 by applying l'Hopital's rule once. Finally, we conclude that the statement holds for all integers i≥0.

Step-by-step explanation:

To prove that lim x→[infinity] xᶦ/eˣ = 0 for all integers i≥0 using induction and l'Hopital's rule, we will start by proving the base case i = 0. In this case, we have x⁰/eˣ which equals 1/eˣ, and as x approaches infinity, eˣ also approaches infinity. Therefore, the limit is 0.

Now, assume that the statement holds for i = k. That is, lim x→[infinity] xᵏ/eˣ = 0. We will prove that it holds for i = k+1. Taking the derivative of both the numerator and denominator, we get (kxᵏ-1)/(eˣ). Applying the limit as x approaches infinity, we get (infinity)/(infinity) which is an indeterminate form. Applying l'Hopital's rule once, we get ((k(k-1)xᵏ-2)/(eˣ)). Again, applying the limit as x approaches infinity, we get 0. Therefore, the statement holds for i = k+1.

Since we have proven the base case and the inductive step, the statement holds for all integers i≥0. Hence, lim x→[infinity] xᶦ/eˣ = 0 for all i≥0.