How to solve this limit without (L'Hôpital's rule) I need to solve this limit without L'Hôpital's rule. These questions always seem to have some algebraic trick which I just can't see this time.
$$ \bbox[yellow]
{
\lim_{x\to 2} \left( \frac{e^2-e^x} {2-x}  \right)
}
$$
Could someone give me a hint as to what I need to do to the fraction to make this work? Thanks!

since $\lim _{ x\to 0 } \frac { { e }^{ x }-1 }{ x } =1$ we have 
$$\lim _{ x\to 2 } \left( \frac { e^{ 2 }-e^{ x } }{ 2-x }  \right) =\lim _{ x\to 2 } \frac { { e }^{ x }\left( { e }^{ 2-x }-1 \right)  }{ 2-x } ={ e }^{ 2 }$$
