# How do you find lim (e^t-1)/t as t->0 using l'Hospital's Rule?

Jun 1, 2018

${\lim}_{t \to 0} \frac{{e}^{t} - 1}{t} = 1$

#### Explanation:

We have

$L = {\lim}_{t \to 0} \frac{{e}^{t} - 1}{t}$

To apply L'Hôpital's rule, we must have a $0 \text{/} 0$ or $\infty \text{/} \infty$ situation. If we plug in $t = 0$ we find that:

$L = \frac{{e}^{0} - 1}{0} = \frac{0}{0}$

Thus we can use L'Hôpital's rule, which states that:

$L = {\lim}_{t \to 0} \frac{{e}^{t} - 1}{t} = \left(\text{d"/("d"t) (e^t-1))/("d"/("d} t\right) t$

We know that ${e}^{x}$ is one of the functions with the property that $f ' \left(x\right) = f \left(x\right)$, and as $- 1$ is just a constant, it will vanish when we take the derivative.

$\therefore L = {\lim}_{t \to 0} \frac{\textcolor{red}{{e}^{t}}}{\textcolor{red}{1}} = {\lim}_{t \to 0} \textcolor{red}{{e}^{t}}$

$\textcolor{red}{L = {e}^{0} = 1}$

Jun 1, 2018

${\lim}_{t \rightarrow 0} \frac{{e}^{t} - 1}{t} = 1$

#### Explanation:

${\lim}_{t \rightarrow 0} \frac{{e}^{t} - 1}{t} {=}_{D L H}^{\left(\frac{0}{0}\right)} {\lim}_{t \rightarrow 0} {e}^{t} = {e}^{0} = 1$