Can we apply L'Hopital's rule? Here is a question I have:
I have to calculate the limit $ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t} }$. Can we apply the L'Hopital rule or I have to write it as: 
$$ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t}  = \lim_{t \to 0} \frac{e^t-1}{t} \cdot \frac{1}{e^t} =1 \cdot 1=1 }$$
Is $ \displaystyle{ \lim_{t \to 0} \frac{e^t-1}{t} }$ a "basic" limit that cannot be calculated using L'Hopital rule?
Thank's in advance!
edit: I was made I typo. Now it is the correct.
Can we apply L'Hopital's rule to calculate the limit $ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t} }$ ?
Sorry for the confusion.
 A: The initial limit you want to calculate is not used in proving that $(e^t)'=e^t$, so you can use l'Hospital. Either way, I think your argument is shorter.
Now I realize that you asked if you can calculate $\lim_{t \to 0} \frac{e^t -1}{t}$ using l'Hospital. You cannot do that. That limit is elementary, and you do not have to prove it every time. You cannot use l'Hospital for this limit because the limit itself is used when proving that $(e^t)'=e^t$. 
A: Your method is essentially equivalent to L'Hospital's rule for functions of this form, viz.
$$\rm \left(\frac{f(x)-f(0)}{x}\right) \left(\frac{1}{f(x)}\right)\ \to\ f'(0)\:\frac{1}{f(0)}$$
$$\rm \frac{(f(x)-f(0))'}{(x\:f(x))'}\ =\ \frac{f'(x)}{f(x)+x\:f'(x)}\ \to\ \frac{f'(0)}{f(0)} $$
A: You can use Lopital,the only thing is that you have a mistake in the derivate 
of the denominator (+ and not -).You can simplify e^t.
But the final limit is 1.
My answer was for your original post.Now I see you changed the post.
A: Observe that when $t$ approaches $0$ we have an indeterminate form that is:
$$\lim_{t\to 0}\frac{e^t-1}{te^t}=\frac{1-1}{0}=\frac{0}{0}$$ So we can apply the L'Hopital rule:(take the derivative of numerator and denominator w.r.t $t$)
$$\lim_{t\to 0}\frac{e^t-1}{te^t}=\lim_{t\to 0}\frac{e^t}{te^t+e^t}=\frac{1}{0+1}=1.$$
