Question

Find the limit, if it exists.
Lim (2x^3+x^2+7)
X->2

Answer 1

Looking Ahead to Calculus: Lessons 1-5 Quick Check Answers

Lesson 1

Q1 B. -21

Q2 D. 4

Q3 D. 4 (yes, again)

Q4 B. lim f(x) = 5 [x→3+] ;

D. lim f(x) = 6 [x→3-] ;

F. lim f(x) [x→3] does not exist...

Lesson 2

Q1 A. lim x = a [x→a] ;

B. lim a = a [x→a];

D. lim x = 5 [x→5]

Q2 D. 16

Q3 B. 10a

Q4 D. 3

Lesson 3

Q1 A. -1

Q2 D. 1

Q3 B. It is discontinuous because there is a value a for which f(a) is not defined.

Q4 D. It is discontinuous because there is a value a such that lim f(x) [x→a] does not exist.

Lesson 4

Q1 D. 3

Q2 A. -13

Q3 C. y = -2x + 6

Q4 C. f'(a) = 8a + 2

Lesson 5

Q1 C. 2

Q2 D. v(t) = 8t - 2

Q3 B. 902 mi./min.

Q4 A. The car is slowing down at a rate of 10 mi./h^2.

Have a great summer

Answer 2

27

Explanation:

In this case, because we have a polynomial where all x-values exist, we can plug the value (x=2) straight into the function:

`\lim_(x \to \ 2) (2x^3+x^2+7)`

`= 2(2)^3+(2)^2+7`

`= 2(8)+4+7`

`= 27`

The limit of (2x^3+x^2+7) as x approaches 2 equals 27, which makes sense since the value of (2x^3+x^2+7) at x = 2 is 27.