MAT135 Calculus 1 Final Exam | Questions
and Answer Key | 2026 Update | 100%
Correct - University of Toronto.
SECTION 1: LIMITS & CONTINUITY
Question 1
Evaluate: limx→+∞2x2−11+x−3x2limx→+∞1+x−3x22x2−1
A. −23−32
B. 2332
C. −13−31
D. Does not exist
,Correct Answer: A. −23−32
Rationale: Divide numerator and denominator
by x2x2: limx→∞2−1/x21/x2+1/x−3=2−3=−23limx→∞1/x2+1/x−32−1/x2=−32=−32.
Question 2
Evaluate: limx→1x2+x−2x2−1limx→1x2−1x2+x−2
A. 0
B. 3223
C. 1
D. Does not exist
Correct Answer: B. 3223
Rationale: Factor numerator: (x+2)(x−1)(x+2)(x−1).
Denominator: (x−1)(x+1)(x−1)(x+1).
Cancel x−1x−1. limx→1x+2x+1=32limx→1x+1x+2=23.
Question 3
Evaluate: limx→0sin3x2xlimx→02xsin3x
A. 0
B. 3223
C. 1221
D. Does not exist
Correct Answer: B. 3223
Rationale: Using limx→0sin3x3x=1limx→03xsin3x
=1. sin3x2x=32⋅sin3x3x→322xsin3x=23⋅3xsin3x→23.
,Question 4
Evaluate: limx→∞sinxxlimx→∞xsinx
A. 1
B. 0
C. Does not exist
D. ∞∞
Correct Answer: B. 0
Rationale: Since −1≤sinx≤1−1≤sinx≤1, we have −1x≤sinxx≤1x−x1≤xsinx≤x1. By
the squeeze theorem, the limit is 0.
Question 5
Evaluate: limx→0xsinxlimx→0xsinx
A. 0
B. 1
C. ∞∞
D. Does not exist
Correct Answer: B. 1
Rationale: Let y=xsinxy=xsinx.
Then lny=sinxlnxlny=sinxlnx. limx→0+sinxlnx=0limx→0+
sinxlnx=0. Thus y→e0=1y→e0=1.
Question 6
, State the limit definition of the derivative of a function f(x)f(x):
A. limh→0f(x+h)−f(x)hlimh→0hf(x+h)−f(x)
B. limh→∞f(x+h)−f(x)hlimh→∞hf(x+h)−f(x)
C. limh→0f(x)−f(x−h)hlimh→0hf(x)−f(x−h)
D. limx→af(x)−f(a)x−alimx→ax−af(x)−f(a)
Correct Answer: A. limh→0f(x+h)−f(x)hlimh→0hf(x+h)−f(x)
Rationale: This is the standard limit definition. Both forms in A and D are equivalent
definitions .
Question 7
State the ε−δε−δ definition of limx→af(x)=Llimx→af(x)=L:
A. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
B. For every δ>0δ>0, there exists ε>0ε>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
C. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣>ε∣f(x)−L∣>ε whenever ∣x−a∣<δ∣x−a∣<δ
D. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣>δ∣x−a∣>δ
Correct Answer: A. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
Rationale: This is the formal definition of a limit at a point using epsilon-delta notation .
Question 8
and Answer Key | 2026 Update | 100%
Correct - University of Toronto.
SECTION 1: LIMITS & CONTINUITY
Question 1
Evaluate: limx→+∞2x2−11+x−3x2limx→+∞1+x−3x22x2−1
A. −23−32
B. 2332
C. −13−31
D. Does not exist
,Correct Answer: A. −23−32
Rationale: Divide numerator and denominator
by x2x2: limx→∞2−1/x21/x2+1/x−3=2−3=−23limx→∞1/x2+1/x−32−1/x2=−32=−32.
Question 2
Evaluate: limx→1x2+x−2x2−1limx→1x2−1x2+x−2
A. 0
B. 3223
C. 1
D. Does not exist
Correct Answer: B. 3223
Rationale: Factor numerator: (x+2)(x−1)(x+2)(x−1).
Denominator: (x−1)(x+1)(x−1)(x+1).
Cancel x−1x−1. limx→1x+2x+1=32limx→1x+1x+2=23.
Question 3
Evaluate: limx→0sin3x2xlimx→02xsin3x
A. 0
B. 3223
C. 1221
D. Does not exist
Correct Answer: B. 3223
Rationale: Using limx→0sin3x3x=1limx→03xsin3x
=1. sin3x2x=32⋅sin3x3x→322xsin3x=23⋅3xsin3x→23.
,Question 4
Evaluate: limx→∞sinxxlimx→∞xsinx
A. 1
B. 0
C. Does not exist
D. ∞∞
Correct Answer: B. 0
Rationale: Since −1≤sinx≤1−1≤sinx≤1, we have −1x≤sinxx≤1x−x1≤xsinx≤x1. By
the squeeze theorem, the limit is 0.
Question 5
Evaluate: limx→0xsinxlimx→0xsinx
A. 0
B. 1
C. ∞∞
D. Does not exist
Correct Answer: B. 1
Rationale: Let y=xsinxy=xsinx.
Then lny=sinxlnxlny=sinxlnx. limx→0+sinxlnx=0limx→0+
sinxlnx=0. Thus y→e0=1y→e0=1.
Question 6
, State the limit definition of the derivative of a function f(x)f(x):
A. limh→0f(x+h)−f(x)hlimh→0hf(x+h)−f(x)
B. limh→∞f(x+h)−f(x)hlimh→∞hf(x+h)−f(x)
C. limh→0f(x)−f(x−h)hlimh→0hf(x)−f(x−h)
D. limx→af(x)−f(a)x−alimx→ax−af(x)−f(a)
Correct Answer: A. limh→0f(x+h)−f(x)hlimh→0hf(x+h)−f(x)
Rationale: This is the standard limit definition. Both forms in A and D are equivalent
definitions .
Question 7
State the ε−δε−δ definition of limx→af(x)=Llimx→af(x)=L:
A. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
B. For every δ>0δ>0, there exists ε>0ε>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
C. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣>ε∣f(x)−L∣>ε whenever ∣x−a∣<δ∣x−a∣<δ
D. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣>δ∣x−a∣>δ
Correct Answer: A. For every ε>0ε>0, there exists δ>0δ>0 such
that ∣f(x)−L∣<ε∣f(x)−L∣<ε whenever ∣x−a∣<δ∣x−a∣<δ
Rationale: This is the formal definition of a limit at a point using epsilon-delta notation .
Question 8
