0

I am unable to prove the following:

$M^{\theta} \sim 1 + \log(M)\theta$ as $\theta \to 0$. Here, $M$ is a constant.

ksank43
  • 97
  • 1
    Since $e^x\sim 1+x$ as $x\to 0$ and $M^\theta=e^{\theta\log M}$, $$M^\theta\sim 1\color{red}{+}\log(M)\theta $$ as $\theta\to 0$. – Jack D'Aurizio Oct 11 '16 at 23:58
  • 1
    Actually, $M^{\theta}-1 \sim \log(M)\theta$ as $\theta \to 0$, which is quite different from $M^{\theta}\sim1+ \log(M)\theta$. – Did Oct 12 '16 at 00:03

0 Answers0