Question

\lim_{x\to 0^+} (-mx)^{\tan x}

Answer 1

1

Answer 2

The limit is of the indeterminate form $0^0$. Taking the natural logarithm transforms it into a form suitable for L'Hopital's Rule. Assuming $m<0$ for the base to be positive in real numbers, or considering complex values for $m>0$, the limit of the logarithm is 0, leading to the original limit being $e^0=1$. The case $m=0$ yields a limit of 0. For a single answer, $m \neq 0$ is implicitly assumed, giving 1.