AP Calculus AB Unit 1 Limits and Continuity Test Study

• -1

• 7

• -2

• 6

Answer explanation

use direct substitution, sub 2 in for x and follow order of operations

Tags

CCSS.HSA.APR.A.1

• 2

• 8

• -2

• 0

Tags

CCSS.HSA.APR.D.6

• -2

• 1

Tags

CCSS.HSF-IF.C.8B

• -1

• -2

• 0

• 2

• 3

• 4

Answer explanation

the limit as x approaches 2 from the left is equal to the limit at x approaches 2 from the right there fore the limit as x approaches 2 is 0. It does not matter that f(2)=2. This is also known as a removable discontinuity. Must be able to write this mathematically $\lim_{x\rightarrow2^-}\ f\left(x\right)\ =\ \lim_{x\rightarrow2^+}\ f\left(x\right)\ =0\ therefore\ \lim_{x\rightarrow2}\ f\left(x\right)=2$

Tags

CCSS.HSA.REI.D.10

CCSS.HSA.REI.D.11

• 0

• ∞

• - ∞

• DNE

Answer explanation

$\lim_{x\rightarrow2^-}\ F\left(x\right)=-\infty$ $\lim_{x\rightarrow2^+}\ f\left(x\right)=\infty$
therefore the limit does not exist (bc they are different)

Tags

CCSS.HSA.REI.D.10

Which of the following best describes the continuity at x = 1?

• Continuous

• Removable Point Discontinuity

• Non-removable Infinite Discontinuity

• Non-removable Jump Discontinuity

Answer explanation

graph the function and look! know your different types of discontinuities

Tags

CCSS.HSF-IF.C.7B

• Continuous

• Removable Point Discontinuity

• Non-removable Infinite Discontinuity

• Non-removable Jump Discontinuity

Answer explanation

because (x-5) is a factor of the numerator and the denominator it is a removable point discontinuity

Tags

CCSS.HSA.APR.D.7

• Does not exist

• -7/5

• -5/9

• -1/2

Answer explanation

Use direct substitution(plug in!)

Tags

CCSS.HSA.APR.D.6

CCSS.HSA.APR.D.7

CCSS.HSA.SSE.B.3

• Does Not Exist

• 9

• 1

• 0

Answer explanation

the degree of the denominator is greater than the degree of the numerator therefore it must approach 0. Must be able to prove algebraically