[L'Hospital's Rule] Can I Use Mathematical Induction to Prove This?

[Argonaut]

Homework Statement

Prove that

$$\lim_{x\to\infty} \frac{e^x}{x^n}=\infty$$

for any positive integer n. This shows that the exponential function approaches infinity faster than any power of x.

[Stewart Calculus, 4.4.73]

Relevant Equations

Derivative of Exponential Functions
Power Rule
L'Hospital's Rule
Mathematical Induction

The Student's Manual simply applies l'Hospital's Rule n-times to arrive at $\frac{e^x}{n!}\to\infty$ as $x\to\infty$.

However, I'm wondering if I could use Mathematical Induction to prove this. Is the following correct and sufficiently rigorous (at least for an undergraduate-level Calculus 1 course)?

Show that the statement is true for n=1:
We apply l'Hospital's Rule.

$$\lim_{x\to\infty} \frac{e^x}{x^1}=\lim_{x\to\infty} e^x = \infty$$

Show that the statement is true for n+1 if it is true for n:
Assume $\lim_{x\to\infty} \frac{e^x}{x^n}=\infty$. Then by applying l'Hospital's Rule to case n+1, we have

$$
\lim_{x\to\infty} \frac{e^x}{x^{n+1}}=
\lim_{x\to\infty} \frac{e^x}{(n+1)x^{n}}=
\frac{1}{n+1}\lim_{x\to\infty} \frac{e^x}{x^{n}}=\infty
$$

Therefore, the statement has been proven true for all positive integers n by the principle of Mathematical Induction.

 

[anuttarasammyak]

Why don't you repeat derivatives. $x^n$ becomes n! after n, which is finite, times derivative. $e^x$ remains same after any times of derivative.

 

[FactChecker]

I think that your proof is valid. In a real proof, you would want to be more precise and say exactly which equality was from an application of L'Hospital's Rule.
Also, notice that you do not have to start the induction from n=1. You could start from n=0 and then you only need to apply L'Hospital's Rule once. (It's a little simpler.)

 

[Argonaut]

Why don't you repeat derivatives. $x^n$ becomes n! after n, which is finite, times derivative. $e^x$ remains same after any times of derivative.

That's the solution the Student Solutions Manual provided.

I think that your proof is valid. In a real proof, you would want to be more precise and say exactly which equality was from an application of L'Hospital's Rule.
Also, notice that you do not have to start the induction from n=1. You could start from n=0 and then you only need to apply L'Hospital's Rule once. (It's a little simpler.)

I see! I started from n=1 because the exercise explicitly said n was "any positive integer".

 

[FactChecker]

I see! I started from n=1 because the exercise explicitly said n was "any positive integer".

I see. So what they asked you to prove is included in what you could prove. Also, notice that the proof for the first one can be much different from the proof for the inductive n+1 step. The first one is often a relatively simple degenerate case.

 

[Argonaut]

I see. So what they asked you to prove is included in what you could prove. Also, notice that the proof for the first one can be much different from the proof for the inductive n+1 step. The first one is often a relatively simple degenerate case.

I've got you - thanks for the feedback!

 

[anuttarasammyak]

Taylor expantion of e^x
e^x=\sum_{k=0}^\infty \frac{x^k}{k!}
\frac{e^x}{x^n}=\sum_{k=0}^\infty \frac{x^{k-n}}{k!}
So we know e^x becomes infinity faster than x^n for any n.