MAT1512
Semester 2 2018
Assignment 01 Memorandum
, MAT1512
Semester 2 2018
Assignment 01 Memorandum
QUESTION 1
(a)
(i) The value of the function 𝑓(𝑥) at 𝑥 = 1 is indicated by the filled
circle at coordinates (1,3). Therefore, 𝑓(1) = 3.
From the graph we see that the values of 𝑓(𝑥) approach 2 as 𝑥
approaches 1 from the left and right respectively. Therefore:
(ii) lim 𝑓 (𝑥) = 2
𝑥→1−
(iii) lim+ 𝑓 (𝑥) = 2
𝑥→1
(iv) Since the left and right limits from (ii) and (iii) above are equal, we
have that:
lim 𝑓(𝑥) = 2
𝑥→1
(v) The value of the function 𝑓(𝑥) at 𝑥 = 3 is indicated by the filled
circle at coordinates (3,2). Therefore, 𝑓(3) = 2.
From the graph we see that the values of 𝑓(𝑥) approach 3 as 𝑥
approaches 3 from the left, but approaches 1 as 𝑥 approaches 3 from the
right. Therefore:
(vi) lim 𝑓 (𝑥) = 3
𝑥→3−
(vii) lim+ 𝑓 (𝑥) = 1
𝑥→3
(viii) Since the left and right limits from (vi) and (vii) above are
different, we have that:
lim 𝑓 (𝑥) does not exist.
𝑥→3
(ix) The value of the function 𝑓(𝑥) at 𝑥 = 2, we read it off of the graph and
we see that 𝑓(2) ≈ 2.5.
From the graph we see that the values of 𝑓(𝑥) approach approximately 2.5
as 𝑥 approaches 2 from the left and right respectively. Therefore:
(x) lim 𝑓 (𝑥) ≈ 2.5
𝑥→2−
(xi) lim 𝑓 (𝑥) ≈ 2.5
𝑥→2+
© 2018
James Training Solutions
Learning and growing together.
Semester 2 2018
Assignment 01 Memorandum
, MAT1512
Semester 2 2018
Assignment 01 Memorandum
QUESTION 1
(a)
(i) The value of the function 𝑓(𝑥) at 𝑥 = 1 is indicated by the filled
circle at coordinates (1,3). Therefore, 𝑓(1) = 3.
From the graph we see that the values of 𝑓(𝑥) approach 2 as 𝑥
approaches 1 from the left and right respectively. Therefore:
(ii) lim 𝑓 (𝑥) = 2
𝑥→1−
(iii) lim+ 𝑓 (𝑥) = 2
𝑥→1
(iv) Since the left and right limits from (ii) and (iii) above are equal, we
have that:
lim 𝑓(𝑥) = 2
𝑥→1
(v) The value of the function 𝑓(𝑥) at 𝑥 = 3 is indicated by the filled
circle at coordinates (3,2). Therefore, 𝑓(3) = 2.
From the graph we see that the values of 𝑓(𝑥) approach 3 as 𝑥
approaches 3 from the left, but approaches 1 as 𝑥 approaches 3 from the
right. Therefore:
(vi) lim 𝑓 (𝑥) = 3
𝑥→3−
(vii) lim+ 𝑓 (𝑥) = 1
𝑥→3
(viii) Since the left and right limits from (vi) and (vii) above are
different, we have that:
lim 𝑓 (𝑥) does not exist.
𝑥→3
(ix) The value of the function 𝑓(𝑥) at 𝑥 = 2, we read it off of the graph and
we see that 𝑓(2) ≈ 2.5.
From the graph we see that the values of 𝑓(𝑥) approach approximately 2.5
as 𝑥 approaches 2 from the left and right respectively. Therefore:
(x) lim 𝑓 (𝑥) ≈ 2.5
𝑥→2−
(xi) lim 𝑓 (𝑥) ≈ 2.5
𝑥→2+
© 2018
James Training Solutions
Learning and growing together.