AP Calculus AB Unit 1 - 1.2-1.6 Review - Limits - Solutions

Sep 30, 2026

AP Calculus AB Unit 1 review solutions focused on limits and continuity. The materials include questions based on a given graph and functions evaluated at specific x-values, followed by multiple limit problems involving one-sided limits, infinite behavior, and cases where limits do or do not exist (including DNE and infinity forms). Additional items apply continuity via the definition at points of discontinuity, and request analysis of removable versus nonremovable discontinuities, vertical asymptotes, and related true/false statements.

LanguageEnglish

CategoryExam

FormatPDF

Pages5

Reading Time13 min

AccessFree to download

Unit 1 (1.2-1.6) Review - Limits - Solutions

Part I: No Calculator (Problems 1-37)

Part II: Calculators Allowed (First Part of Section)

AP Calculus AB Name: Solutions Unit 1 (1.2-1.6) Review-LIMITS PART I- DO NOT USE A CALCULATOR ON ANY PROBLEM IN THIS SECTION. (Problems 1-37) Consider the graph of function, f , shown below. | | | | | | | | | | | | 6 | | | | | | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | | | | | | | | | | | | | | | | | | f | | | | | | | | | | | | | | | | 4 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 2 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | -1 | 0 | | | | - | 5 | | | | | | | | | 5 | | | | | 1 |
0 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | -2 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | -4 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Answer the following questions about function f. 1.) f (-5) = 6 2.) f (2) = 1 3.) f (4) = -3 4.) lim f (x) = 2 5.) lim f (x) = 6 6.) limf (x) = 3 x®-7 x®-5 x®2 7.) limf (x) = DNE 8.) limf (x) = DNE 9.) lim f (x) = -¥DNEx®4 x®0 - x®0 10.) lim f (x) = +¥ DNE 11.) lim f (x) = -3 12.) lim f (x) = 2 x®0+ x®4+ x®4- 13.) lim f (x) =0+ 14.) limf (x) =0- x®-¥ x®¥ 15.) Use the definition of a continuous function at a number to answer the following. a. f is not continuous at x = -7 because:__f (-7 ) is not defined b. f is not continuous at x = 2 because: f (2 )¹ limx®2 f (x ) c. f is not continuous at x = 4 because: limf (x) = DNE x®4 DO NOT USE A CALCULATOR | 16.) limx®2 (-x2 + 4x)
= (- (2 )2 + 4(2 ))= 4 | | x-3 ìï9-3 17.) lim ~~ ~~ í
x®9- x-9 ïî9-9 = Do some algebra
= lim x®9-
x
é ùê ~~ ~~ ~~ ~~ úêë ûú

-3 x-9
x

+3 x

+3 | = 0 0 | x ì0
18.) lim ~~ ~~ í x®0 tan x îtan0 = 0
Do some algebra
= limx®0 êéësixnx~~ ~~cos1x~~ ~~ ùûú= (1)(1)= 1 | | --- | --- | --- | --- | --- | | | | = lim x®9-
éêêêë
ù x-9 ú(x-9 )( x +3)ûú
= lim x®9-
é 1 ù
ê ú
ë x +3 û
=
1 6-
| | | | 19.) lim x→−2+ | 􀂧 x 􀂷􀂨 ~~ ~~ 􀂸􀂩 x+ 2 􀂹 | 􀂧 1 􀂷
20.) lx0 − 􀂩􀂨 1+ x 􀂹􀂸
æ 1 ö
= çè1 + 0- ø÷ = -¥ DNE | | 21.) lixm®1 (sinpx )
= (sinp(1))= 0 | | æ -2+ ö -2+
= èç -2 + + 2 ø÷ = 0 + = -¥ DNE | | | | | | 7 − 6x5
22.) lim ~~ ~~x→􀁦 x+ 3
= lxim®¥ ~~ ~~-6xx5 = lxim®¥ (-6x4 ) = -¥ DNE | | 6-t 3 23.) tli®m- ¥ 7t3 +3
-t3 1
= lim ~~ ~~ = - ~~ ~~t®-¥ 7t3 7 | | 24.) lim x-2 x®-¥ x 2 + 2x +1
x 1
= lim ~~ ~~ = lim ~~ ~~x®-¥ x 2 + 2x x®-¥ x + 2 1 -
= = 0 -¥ | | 2 − y
25.) lim ~~ ~~y→−􀁦 7 + 6y2
−y −y
= lim ~~ ~~ = lim ~~ ~~y→−􀁦 6y2 y→−􀁦 y 6 1
=
6 | | 26.) 2 f (x) when
􀂭 x2 − 3x+ 6, x 􀀟 2 f (x) = 􀂮 2
􀂯−x + 3x+ 2, x 􀁴 2 lxim®2-f (x )= 22-3(2 )+6 = 4 lxim®2+ f (x )= -22 +3(2 )+ 2 = 4 limx®2 f (x )= 4 | | x2 − a2
27.) If a 􀁺 0, then lxa x4 − a4 is:
x 2-a2 ìï (-a )2-a2 = 0
lim í
x® -a x 4-a4 îï(-a )4-a4 = 0
Do some algebra
x 2-a2 lxim®-a (x 2-a2 )(x 2 + a2 )
1 1
= lim ~~ ~~ = ~~ ~~x®-a 1 (x 2 + a2 ) 2a2 | | 28.) Find a c such that f (x) is continuous on the entire real line. | | | 29.) Find the x-values (if any) at which f is discontinuous. Label as removable or nonremovable.
f (x )= 22xx2~~ ~~+-618
f (x )= 22xx2~~ ~~~~ ~~618 = xx2~~ ~~+-39 x +3 1
= ~~ ~~ = ~~ ~~
(x +3)(x-3) x-3 f (x ) is not defined at x = -3 xli®m-3+ f (x )= xli®m-3+ f (x )= - ~~ ~~16 removable discontinuity at x = -3 because the limit exists but the function value does not. | | | 30.) Determine all of the vertical asymptotes of f(x):
f (x) = xx2~~ ~~24
x + 2 1

FAQs

What topics are covered in this Unit 1 review?

The review focuses on evaluating limits and applying the definition of continuity, including removable and nonremovable discontinuities, vertical asymptotes, and when limits do not exist (DNE).

When are calculators allowed in the solutions set?

Calculators are not allowed for Part I (Problems 1-37). Calculators may be used on the first part of Part II of the section.

How does the document handle points where the function is undefined or discontinuous?

It distinguishes cases where the limit exists but the function value does not (removable discontinuity) and cases involving one-sided limits or where the limit does not exist (DNE), using specific x-values and continuity checks.

Sep 30, 2026

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