# What is the limit of (3+h)^-1 -3^(-1/h) as h approaches 0?

Sep 10, 2016

${\lim}_{h \to {0}^{-}} {\left(3 + h\right)}^{-} 1 - {3}^{- \frac{1}{h}} = - \infty$

${\lim}_{h \to {0}^{+}} {\left(3 + h\right)}^{-} 1 - {3}^{- \frac{1}{h}} = \frac{1}{3}$

so the limit DNE

#### Explanation:

${\lim}_{h \to 0} {\left(3 + h\right)}^{-} 1 - {3}^{- \frac{1}{h}}$

$= {\lim}_{h \to 0} \textcolor{b l u e}{\frac{1}{3 + h}} - \textcolor{red}{\frac{1}{3} ^ \left(\frac{1}{h}\right)}$

the limit of the sum is the sum of the limits

$= {\lim}_{h \to 0} \textcolor{b l u e}{\frac{1}{3 + h}} - {\lim}_{h \to 0} \textcolor{red}{\frac{1}{3} ^ \left(\frac{1}{h}\right)}$

$= \frac{1}{3} - {\lim}_{h \to 0} \textcolor{red}{\frac{1}{3} ^ \left(\frac{1}{h}\right)}$

the blue term clearly tends to $\frac{1}{3}$ for all $0 < \left\mid h \right\mid \text{ << } 1$, so we are left to consider the red term

$L = - {\lim}_{h \to 0} {3}^{- \frac{1}{h}}$

$L = - {3}^{{\lim}_{h \to 0} - \frac{1}{h}}$ as the exponential function is continuous

the sign of the exponent will be different coming at this from either side

$- {3}^{{\lim}_{h \to {0}^{-}} - \frac{1}{h}} = - {3}^{\infty} = - \infty$

$- {3}^{{\lim}_{h \to {0}^{+}} - \frac{1}{h}} = - {3}^{- \infty} = 0$