Under what conditions is $\lim f(x)=e^{\lim \ln(f(x))}$

Question

Under what conditions is $\lim_{x\to c} f(x)=e^{\lim_{x \to c} \ln(f(x))}$?

I saw this limit in an article used to show that:

$$\lim_{\rho \to 0} [\alpha x_{1}^{\rho} + (1-\alpha) x_{2}^{\rho}]^{\frac{1}{\rho}} = x_{1}^{\alpha}x_{2}^{1-\alpha}$$

where the "trick" is used to apply l'hopitals rule to:

$$\lim_{\rho \to 0} \frac{\ln(\alpha x_{1}^{\rho} + (1-\alpha) x_{2}^{\rho})}{\rho} $$

However I was unaware that this trick existed before today, and I am wondering if there are conditions under which it applies?

Hint: What do you know about the composition of continuous functions? — Zestylemonzi, Sep 26 '15 at 17:24
A sufficient condition is that the limit of $f(x)$ is a positive number. — Simon S, Sep 26 '15 at 17:25
$\lim e^{g(x)} = e^{\lim g(x)}$ whenever either limit exists in $[-\infty,\infty].$ — zhw., Sep 26 '15 at 17:28

score 6 · Accepted Answer · edited Sep 26 '15 at 17:40

The function $e^{x}$ is continuous on $\mathbb{R}$, which means that $$\lim_{x\rightarrow c}e^{x}=e^{\lim_{x\rightarrow c}x}$$ for all $c\in \mathbb{R}$. So, $$e^{\lim_{x\rightarrow c}\ln(f(x))}=\lim_{x\rightarrow c}e^{\ln(f(x))}$$ always. Now, for all a such that $\ln(a)$ is defined (i.e., $a>0$), it holds that $e^{\ln(a)}=a$, so really your question depends mostly on the range and other properties of the function $f(x)$. Does f take on negative values, or is it true that $f(x)>0$ for all x sufficiently close to c? Does the limit $\lim_{x\rightarrow c}f(x)$ actually exist?

Under what conditions is $\lim f(x)=e^{\lim \ln(f(x))}$

1 Answers1