Answer
$1$
Work Step by Step
Consider $f(t)=\lim\limits_{t \to 0^{+}}\dfrac{e^t}{t}-\dfrac{1}{t}=\lim\limits_{t \to 0^{+}} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Simplify further:
$\lim\limits_{t \to 0^{+}}\dfrac{e^t}{t}-\dfrac{1}{t}=\lim\limits_{t \to 0^{+}}\dfrac{e^t-1}{t}$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L'Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{t \to 0^{+}}\dfrac{e^t}{1}=1$
