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MA103 Lab Report 1 - Limits
Name:
Student Number:
Winter 2019
1. [4 marks] On the axes provided, sketch a function f (x) on the interval [−5, 5] that satisfies all of the given conditions.
(Many solutions are possible.)
(a) lim− f (x) = 3 , lim+ f (x) = −1, f (2) = −1
x→2
x→2
(b) lim f (x) = 0, lim− f (x) = 1, lim+ f (x) = 2,
x→−3
x→0
x→0
f (−3) = 2, and f (0) = 0
2. [7 marks] Consider the function f (x) =

10(2 − x)



 |x2 + x − 6|




√
if x < 2
1
x−1
if x ≥ 2
(a) Define f in Maple and plot the function for x ∈ [−2, 5].
f(x):=piecewise(x<2,(10*(2−x)/abs((x∧2)+x−6)),x>=2,(1/sqrt(x−1)))
plot(f(x),x=−2..5,discont=true)
(b) Based on the plot, state the value of each of the following limits:
lim f (x) = 1
lim f (x) = 2
x→2−
x→2+
(c) Verify your answer to (b) by evaluating lim f (x) and lim f (x) by hand. Use proper form and show all of your
x→2−
x→2+
work.
lim f (x)
x→2−
= lim−
10(2 − x)
|x2 + x − 6|
= lim
10(2 − x)
|(x − 2) (x + 3)|
=√
= lim
10(2 − x)
because x ∈ (−3, 2)
− (x − 2) (x + 3)
=1
= lim−
10
10
=
=2
(x + 3)
5
x→2
x→2−
x→2−
x→2
lim f (x)
x→2+
(d) Does lim f (x) exist? If so, determine the limit. If not, explain why.
x→2
Since lim− f (x) 6= lim+ f (x), lim f (x) does not exist.
x→2
x→2
x→2
= lim+ √
x→2
1
2−1
1
x−1
3. [4 marks] Evaluate each of the following limits. Show all of your work.
1
1
ex
e4
(b) lim
+
(∞ + ∞)
(a) lim−
u→0 u2 − u
u
0−
x→4 (x − 4)5
1
1
= −∞ since e4 > 0.
+
= lim
u→0 u(u − 1)
u
= lim
1 + (u − 1)
u(u − 1)
= lim
1
= −1
u−1
u→0
u→0
√
4. [5 marks] Consider the function f (x) =
ax + b − 2
, for a, b ∈ R, a 6= 0 (i.e., a and b are constants with a 6= 0).
x
(a) Find the value b such that lim f (x) exists.
x→0
√
lim f (x)
x→0
= lim
x→0
√
= lim
x→0
= lim
x→0
ax + b − 2
x
√
ax + b − 2
ax + b + 2
·√
x
ax + b + 2
ax + b − 4
√
x( ax + b + 2)
√
As x → 0, x( ax + b + 2) → 0. The limit will only exist if the numerator also approaches 0 as x → 0. So, we
must have a(0) + b − 4 = 0 ⇒ b = 4.
(b) Using the value of b determined in (a), find the value a such that lim f (x) = 1.
x→0
lim f (x) = 1
x→0
ax + 4 − 4
√
=1
x→0 x( ax + 4 + 2)
a
=1
⇒ lim √
x→0 ( ax + 4 + 2)
a
⇒ p
=1
( a(0) + 4 + 2)
√
⇒a= 4+2=4
⇒ lim
5. [4 marks] Suppose that |f (x) − ex | < x2 . Use the Squeeze Theorem to determine lim f (x).
x→0
|f (x) − ex | < x2 ⇒ −x2 ≤ f (x) − ex ≤ x2 ⇒ ex − x2 ≤ f (x) ≤ x2 + ex
By the Squeeze Theorem, we have
lim (ex − x2 ) ≤ lim f (x) ≤ lim (x2 + ex )
x→0
x→0
x→0
⇒ e0 − 0 ≤ lim f (x) ≤ 0 + e0
x→0
⇒ 1 ≤ lim f (x) ≤ 1
x→0
∴ lim f (x) = 1
x→0
Grade:
24
24