#### To determine

**a)To calculate:** The limit limt→∞v, where v=mgc(1−e−ct/m)

#### Answer

limt→∞v=mgc

#### Explanation

We compute the limit as t→∞

limt→∞v=limt→∞mgc(1−e−ct/m)=mgclimt→∞(1−e−ct/m)=mgc(1−e−∞)=mgc(1−0)=mgc

We Know that as x→∞⇒e−x→0

**Final statement:** We can conclude that as time increases velocity of the object depends on its mass, acceleration due to gravity and gravitational constant

limt→∞v=mgc

#### To determine

**b)To calculate:**For fixed t, calculate limc→0+v

#### Explanation

limc→0+v=limc→0+mgc(1−e−ct/m)=mglimc→0+1c(1−e−ct/m) (00 Form)Applying l'Hospital's Rule, we get=mglimc→0+−e−ct/m(−t/m)1=mg(t/m)=gt

We can conclude that in a vacuum a falling object would continue to accelerate and its velocity will increase indefinitely

**Final statement:** v=gt