If ƒ and g are differentiable and lim x → c f ( x ) = lim x → c g ( x ) = 0 {\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0\,} or ± ∞ {\displaystyle \pm \infty }
and lim x → c f ′ ( x ) / g ′ ( x ) {\displaystyle \lim _{x\to c}f'(x)/g'(x)} exists,
then lim x → c f ( x ) g ( x ) = lim x → c f ′ ( x ) g ′ ( x ) . {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}.}
Find lim t → 0 exp 3 t − 1 t {\displaystyle \lim _{t\to 0}{\frac {\exp {3t}-1}{t}}}
lim t → 0 exp 3 t − 1 = 0 {\displaystyle \lim _{t\to 0}\exp {3t}-1=0}
lim t → 0 t = 0 {\displaystyle \lim _{t\to 0}t=0}
f ( t ) = exp 3 t − 1 {\displaystyle f(t)=\exp {3t}-1}
f ′ ( t ) = 3 e x p 3 t {\displaystyle f'(t)=3exp{3t}}
g ( t ) = t {\displaystyle g(t)=t} g ′ ( t ) = 1 {\displaystyle g'(t)=1}
l i m t − > 0 ( e ( 3 t ) − 1 ) / t = l i m t − > 0 ( 3 e ( 3 t ) ) / 1 = l i m t − > 03 e ( 3 t ) = 3 {\displaystyle limt->0(e^{(}3t)-1)/t=limt->0(3e^{(}3t))/1=limt->03e^{(}3t)=3}
or 
exists,
