Limit of a log function using expansion of $\log (1+x)$

Question

I am trying to find the limit of

$$\lim_{n\to \infty} n\log \left(1+\left(\frac{f(x)}{n}\right)^\alpha\right)$$

where $f$ is some function $f:X\to[0,\infty]$ and $1\leq \alpha < \infty$ in some other problem I am looking at.

I cannot use the expansion $\log (1+x)=x-\frac{x^2}{2}+\cdots$ because we don't know if $|x|<1$. I know that the answer is $f(x)$ for $\alpha=1$ and $0$ for $\alpha>1$ but I'm not sure how to show it?

For any fixed $x$ and fixed $\alpha\ge1$, there exists an $N$ such that whenever $n>N$, $\left(\frac{f(x)}{n}\right)^\alpha<1$. — Mark Viola, Oct 29 '20 at 03:09

score 2 · Accepted Answer · answered Oct 29 '20 at 03:14

2

HINT:

For any fixed $x$ and fixed $\alpha\ge1$, there exists an $N$ such that whenever $n>N$, $\left(\frac{f(x)}{n}\right)^\alpha<1$.

Then note that for $t\ge0$ we have

$$\log(1+t)=t+O(t^2)$$

Can you finish?

answered Oct 29 '20 at 03:14

Mark Viola

179,405

score 2 · Answer 2 · answered Oct 29 '20 at 03:16

$|\frac{f(x)}{n}|\lt 1$ for fixed $x$ and sufficiently large $n$, so you use the approximation for the log ~ $(\frac{f(x)}{n})^\alpha$. For large n the expression is then ~ $\frac{f(x)^\alpha}{n^{\alpha-1}} \to 0$, as $n\to \infty$ for $\alpha \gt 1$.

Limit of a log function using expansion of $\log (1+x)$

2 Answers2