#### To determine

**To evaluate:** the limit limx→0(f(x)g(x)) and limx→0(f′(x)g′(x)) where f(x)=ex−1 ,g(x)=x3+4x

#### Answer

The value of the limit is limx→0(f(x)g(x))=limx→0(ex−1x3+4x)is undefined

limx→0(f′(x)g′(x))=limx→0(ex3x2+4)=0.25

#### Explanation

From the graph we can make out that the value of the limit limx→0(ex−1x3+4x) is undefined

The blue line indicates (ex−1x3+4x) and the red line indicates (ex3x2+4)

Exact value of the limit is given by

limx→0(ex−1x3+4x) (00Form)Applying l'Hospital's Rulelimx→0(ex−1x3+4x)=limx→0(ex3x2+4)=e04=14=0.25

From the graph we can make out that the value of the limit limx→0(ex3x2+4) is 0.25

The blue line indicates f(x)=ex and the black line indicates g(x)=3x2+4

**Conclusion:** The value of the limits are limx→0(ex−1x3+4x)=0 and limx→0(ex3x2+4)=0.25