explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.
Thanks.
$\displaystyle \lim_{x\to 0^{+}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{+}}\frac{1}{x}\log(2)=\infty $
$\displaystyle \lim_{x\to 0^{-}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{-}}\frac{1}{x}\log(2)=-\infty $
Remember,
$\displaystyle e^{\infty }=\infty$ and $\displaystyle e^{-\infty }=0$
$\displaystyle x=1,\ 2^{\frac{1}{x}}=2^1=2$
$\displaystyle x=0.5,\ 2^{\frac{2(0.5)}{0.5}}=2^2$
$\displaystyle x=0.1,\ 2^{\frac{10(0.1)}{0.1}}=2^{10}$
$\displaystyle As\ x\ goes\ to\ zero\ from\ above,\ f(x)\ goes\ to\ infinity.$
$\displaystyle x=-1,\ f(-1)=2^{-1}=\frac{1}{2}$
$\displaystyle x=-0.5,\ f(-0.5)=2^{-2}=\frac{1}{2^2}$
$\displaystyle x=-0.1,\ f(-0.1)=2^{-10}=\frac{1}{2^{10}}$
$\displaystyle As\ x\ goes\ to\ zero\ from\ below,\ f(x)\ goes\ to\ zero.$