# How do you find the limit of 1-2^-(1/x)/(1+2^-(1/x)) as x approaches 0^-?

Apr 13, 2018

${\lim}_{x \to {0}^{-}} \left[1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)}\right] = 0$

#### Explanation:

We can write $1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)}$

as $1 - \frac{1}{{2}^{\frac{1}{x}} + 1}$ (multiplying numertaor and denominator by ${2}^{\frac{1}{x}}$)

When $x \to {0}^{-}$, ${2}^{\frac{1}{x}} \to 0$

Hence as $x \to {0}^{-}$

$1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)} \to 1 - \frac{1}{1 + 0} = 0$

Hence ${\lim}_{x \to {0}^{-}} \left[1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)}\right] = 0$

Observe that as $x \to {0}^{+}$, as $1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)} = 1 - \frac{1}{{2}^{\frac{1}{x}} + 1}$

as ${2}^{\frac{1}{x}} \to \infty$, $1 - {2}^{-} \frac{\frac{1}{x}}{1 + {2}^{-} \left(\frac{1}{x}\right)} \to 1 - 0 = 1$

graph{1-2^-(1/x)/(1+2^-(1/x)) [-2.5, 2.5, -1.25, 1.25]}