Proof by induction: L'Hospital rule doesn't work, what next?

Question

Show through induction: $\lim_{x\to\infty} \frac{\ln^k x}x=0$

Well the base case of 1 is rather trivial, but when substituting k for k+1 for the inductive hypothesis step we get,

$\lim_{x\to\infty} \frac{\ln^{k+1} x}x=0$

Applying L'Hospital's rule gets is not helpful as we will get

$\lim_{x\to\infty} \frac{(k+1)\ln^{k} x}x$

Applying l 'hospital's again gets:

$\lim_{x\to\infty} \frac{-(k+1)(\ln^{k-1} x) ({\ln (x)} -k)}{x^2} $

Applying l 'hospital's rule runs into an issue of always having an infinity/infinity issue.

Sometimes when l'hospital's rule gets stuck, there is an algebraic manipulation that can be done to overcome the problem. I am not seeing anything. I am wonder what would be another course of action to prove the k+1 case of induction.

Answer 1

I think you might be mistaken in your application of L'Hospital's rule. Assume that case $k$ is true. That is, $$\lim_{x\rightarrow \infty} \frac{\ln^k x}{x}=0.$$ The case $k+1$ is $$\lim_{x\rightarrow \infty} \frac{\ln^{k+1} x}{x}=0,$$ which is of the form $\infty/\infty$.

Apply L'Hospital's rule, $$\lim_{x\rightarrow \infty} \frac{(k+1)\ln^{k} x (1/x)}{1}=(k+1)\lim_{x\rightarrow \infty} \frac{\ln^{k} x}{x}=0,$$ from case $k$.

Answer 2

My diagnosis of the OP's difficulty is not yet being comfortable with mathematical induction. Induction tells us that if we have any statement $P(k)$ about positive integers $k$, and we can prove both

• $P(1)$ is true, and
• for all positive integers $m$, if $P(m)$ is true then $P(m+1)$ is true;

then that comprises a proof that $P(k)$ is true for all positive integers $k$.

In our case, let $P(k)$ be the statement: $\displaystyle\lim_{x\to\infty} \frac{\ln^k x}x = 0$.

• The OP claims that proving $P(1)$ is done, and indeed it follows directly from an application of l'Hopital's rule.
• We now need to prove that for all positive integers $m$, if $P(m)$ is true then $P(m+1)$ is true. Notice that this task is not proving that $P(m)$ is true, nor is it proving that $P(m+1)$ is true. It is literally proving that the implication $P(m) \implies P(m+1)$ is true. And indeed, L'Hopital's rule can be used to show that if $\displaystyle\lim_{x\to\infty} \frac{\ln^m x}x = 0$, then $\displaystyle\lim_{x\to\infty} \frac{\ln^{m+1} x}x = 0$; indeed, this is what was done in manofwar's answer.

Together, these two steps prove (by mathematical induction) that $\displaystyle\lim_{x\to\infty} \frac{\ln^k x}x = 0$ is true for all positive integers $k$.

It's perfectly reasonable to be confused by induction at first, to feel that it's circular or incomplete, or to wonder why proving both the base case and the induction implication suffices to prove the whole universal statement. That's all ok! But I do want to emphasize that this really is a complete proof (and, in particular, that manofwar's answer does successfully overcome the difficulty stated by the OP).

Answer 3

Induction start/inition:

$k=0$: $\lim_{→∞}\frac{\ln^{0}}{}=\lim_{→∞}\frac{1}{}=0$

or

$k=1$: $\lim_{x\to\infty}\frac{\ln^{1}x}{x}=\lim_{x\to\infty}\frac{\ln x}{x}$

This is the case l'Hospital $\frac{\infty}{\infty}$!

So $\lim_{x\to\infty}\frac{\ln^{1}x}{x}\approx_{x \to \infty}\frac{\frac{1}{}}{1}=\lim_{x\to\infty}\frac{1}{x}=0$

This induction start already shows some part of the induction step. The start for $k=1$ makes use of the one for $k=0$.

In words the derivation of the nominator is $\frac{1}{x}$, the one of the denominator is $1$. The limit exists and is $0$.

Induction step:

$k$ is some arbitrary integer. We assume that for this chosen $k$ the identity for the limit to $\infty$ is true.

We now look at $k+1$ following exactly in the Integers the $k$ for which we know the identity holds.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}$

This is indeed a case for the rules of l'Hospital.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}\approx_{x \to \infty}\frac{\infty}{\infty}$

It if allowed to differentiate nominator and denominator separate. The derivative of the nominator is $(k+1)\ln^{k}x\frac{1}{x}$. This is chain rules applied twice. First the outer polynomial of the degree k+1 is derivated, then the in Logarithm is derivated. The derivative of the denominator is $1$.

This quotient is the $k+1$ of the quotient of the induction step premise.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=(k+1)\lim_{x \to \infty}\frac{\ln^{k}x}{x}$

$k$ in integer and fixed so:

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=0$.

For $k=1$ this is simply the extented induction start.

There is nothing more necessary then the derivation of polynomials without coefficients of integer degree and $0$ and the derivation of the $\ln$ (natural logarithm) and the rules of l'Hospital for the case $\frac{\infty}{\infty}$ (quotient of infinity by infinity both positive).

The answer presents the indcution schema following Mathematical_induction. That is the biggest ambiguation overall. I did write explicitly that the induction step premise did assump that the equation, limit holds for all other natural $k$ up to the selected $k$ for the induction step already and that everyone was able to convince him- or herself that this is true.

I have not specially not written that the last equation of manofwar holds. I already factored and wrote the limit identity. This holds only in the limit of infinity:

$$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=(k+1)\lim_{x \to \infty}\frac{\ln^{k}x}{x}$$

Since $k+1$ is a constant for the limit this is true for all integer $k$ after applied the rule of l'Hospital and calculating rules for limits (Limit of a function section Properties).

I did not restrict the presentation to reals.