# How do you find the limit of  (5^t-3^t)/t as t approaches 0?

Jun 27, 2018

$\ln \left(\frac{5}{3}\right)$.

#### Explanation:

Prerequisite : ${\lim}_{h \to 0} \frac{{a}^{h} - 1}{h} = \ln a , \left(a > 0\right)$.

$\text{The Limit} = {\lim}_{t \to 0} \frac{{5}^{t} - {3}^{t}}{t}$,

$= {\lim}_{t \to 0} \frac{\left({5}^{t} - 1\right) - \left({3}^{t} - 1\right)}{t}$

$= {\lim}_{t \to 0} \left\{\frac{{5}^{t} - 1}{t} - \frac{{3}^{t} - 1}{t}\right\}$,

$= \ln 5 - \ln 3$,

$= \ln \left(\frac{5}{3}\right)$.

Jun 27, 2018

${\lim}_{t \to 0} \frac{{5}^{t} - {3}^{t}}{t} = \ln \left(\frac{5}{3}\right)$

#### Explanation:

We know that,

color(red)((1)lim_(hto0)(a^h-1)/h=lna ,where, ain RR^+ -{1}

Let ,

$L = {\lim}_{t \to 0} \frac{{5}^{t} - {3}^{t}}{t}$

$L = {\lim}_{t \to 0} \left[\frac{{5}^{t} \textcolor{b l u e}{- 1} - {3}^{t} \textcolor{b l u e}{+ 1}}{t}\right]$

L=lim_(t to0)(5^t-1)/t-lim_(t to0)(3^t-1)/t...tocolor(red)(Apply(1)

$L = \ln 5 - \ln 3 \to \left[U s e : \ln A - \ln B = \ln \left(\frac{A}{B}\right)\right]$

$L = \ln \left(\frac{5}{3}\right)$