1

I've already seen a couple of times both on questions here (like Value of $\lim_{n\to\infty}{(1+\frac{2n^2+\cos{n}}{n^3+n})^n}$ or Problem of limit of power function) and in other online resources people using the fact that if $f(x), g(x) $ are functions, and $$\lim_{x\to a} f(x) =a$$ $$\lim_{x\to a} g(x) =b$$

Then $$\lim_{x\to a} f(x) ^{g(x)} =a^b $$

But I wasn't able to prove it myself. Any insight on why (and when exactly) is it true?

Community
  • 1
Nescio
  • 2,426

1 Answers1

2

The final claim is not true in general. To begin with, we must make ourselves clear what $x^y$ actually is.

  • If $x,y\in\mathbb N_0$ we can define $x^y$ as the number of maps from a set of $y$ elements to a set of $x$ elements. Specifically, this implies $x^0=1$ for all $x\in\mathbb N_0$ and $x^{y+1}=x\cdot x^y$ for all $x,y\in\mathbb N_0$
  • If $x\in\mathbb R$, $y\in\mathbb N_0$ we can generalize the preceeding and define inductively $x^0=1$, $x^{y+1}=x\cdot x^y$.
  • If $x\in\mathbb R\setminus\{0\}$, $y\in -\mathbb N$, we can define $x^y=\frac1{x^{-y}}$ where the power on the right is defined per the preceeding point. note that we must exclude $x=0$ here.
  • If $x\in\mathbb R_{>0}$, $y\in\mathbb R$ we can define $x^y$ as $\exp(y\ln x)$. Note that this coincides with the preceeding definitions for all applicable cases, i.e., when $x\in\mathbb R_{>0}$ and $y\in\mathbb Z$, but it does not attempt to define $0^y$
  • we can consider a few more special cases but at least when we want to define $x^y$ for arbitrary negative real $x$ and arbitrary real $y$, we run into trouble.

With this in mind, when does $\lim_{x\to c}f(x)^{g(x)}$ even make sense (assuming $\lim_{x\to c}f(x)=a$ and $\lim_{x\to c} g(x)=b$)? First of all, we need the expression to be defined for $x$ sufficiently close to $c$. By the above, we run into trouble for example if $f(x)<0$ and $g(x)\in\mathbb R\setminus\mathbb Z$. note that this may happen even if $a^b$ is defined. For example (with $c=0$) consider $f(x)=-1$, $g(x)=x$; here $\lim f(x)^{g(x)}$ does not exist because the expression is rarely defined; yet $(-1)^0=1$.

The next problem occurs when $a=b=0$. While the expression $0^0$ is defined and has value $1$, the form $0^0$ is indeterminate which means precisely the problem we have: $f(x)\to 0 $ and $g(x)\to 0$ does not imply that $f(x)^{g(x)}\to 1$. You can make up your own collection of examples where $f(x)^{g(x)}$ is not defined often enough; or converges to an arbitrary value; or diverges to infinity; or is divergent because it oscillates.

However, we do have the following:

If $f(x)\to a>0$ and $g(x)\to b$ then $f(x)^{g(x)}\to a^b$. This follows because we can (in fact: must) apply the definition via exponential and logarithm and these two functions are continuous in their domain.

Hagen von Eitzen
  • 374,180