F(x)=2x*e^2x. find lim of f(x) as x--> - infinity and +infinity

I am assuming it is f(x)=2x e^2x 
2x times e raised to the 2x power, and not 2 times e raised to the 2x** 

as it approaches infinity, your f(x) --> infinity 
we can see by inspection, as 2x will approach inf, and e^(2x) will approach infinity even faster. 
inf x inf = inf 

as it approaches neg infinity, your f(x) --> 0 
we can see by inspection, as 2x will approach neg inf, and e^(2x) will approach zero. 
- inf x 0 = - 0 

The minimum(or maximum) is given when the derivative = 0 
Using the product rule, 
f ' (x) = 2e^(2x) + 4x e(2x) 
f ' (x) = 2e^(2x) ( 1 + 2x ) 
We find the roots 
2e^(2x) will never equal 0 
1 + 2x = 0, x = -1/2 

The minimum value will be when x = -0.5

The answer is 1 because you have to flip the sign when dividing by negative number and 3

Provide an equation