Name: Althea Marie R. Jalmanzar
Date: September 18, 2022
Score:____________
WORKSHEET 1- Limit of a Function
A. Create a table of values for the function and use the result to estimate the limit.
1.
sin x
=1
x
x→ 0
lim
X
f(x)
-3
0.047
2.
-2
0.455
-1
0.841
0
undefined
1
0.841
2
0.455
3
0.047
cos x − 1
=0
x
x→ 0
lim
X
f(x)
-3
0.218
(
-2
0.495
-1
0.416
0
undefined
1
1
2
0.270
3
−0.139
)
3. lim x 2 + 1 = 2
x→ 1
X
f(x)
-0.5
0.75
0
1
0.5
1.25
1
2
1.5
3.25
B. Sketch the graph of the function to estimate its limit.
1. 𝐥𝐢𝐦 𝒇(𝒙), where 𝒇(𝒙) = 𝒙𝟐 + 𝟐= 6
𝒙→𝟐
X
0.5
1
1.5
2
2.5
3
3.5
f(x)
2.25
3
4.25
6
8.25
11
14.25
2
5
2.5
7.5
2. lim f ( x ) , where
x→ 1

 x 2 + 3, x  1
f (x) = 

, x=1
2
𝐥𝐢𝐦 𝒙𝟐 + 𝟑 = (𝟏)𝟐 + 𝟑 = 𝟒
𝒙→𝟏
𝐥𝐢𝐦 𝟐 = 𝟐
𝒙→𝟏
X
-0.5
0
0.5
1
1.5
2
2.5
f(x)
3.25
3
3.25
4
5.25
7
9.25
C. Use the graph to find the limit (if it exists). If the limit does not exist, explain why.
1.
lim cos
x→ 0
1
x
The graph showed above to find
the limit does not exist (DNE)
since the lines are too wavy and
compacted to each other. It shows
an oscillating function. The limit
of an oscillating function f(x) as x
approaches positive or negative
infinity is undefined.
2.
4 − x , x  2
f ( x ) , where f ( x ) = 
, x=2
x→ 2
0
lim
𝐥𝐢𝐦 𝟒 − 𝒙 = 𝟒 − 𝟐 = 𝟐
𝒙→𝟐
𝐥𝐢𝐦 𝟎 = 𝟎
𝒙→𝟐
The graph shows a removable
discontinuity. Removable
discontinuity happens when the
two-sided limit exists, but are not
equal to the function's value.
D. Evaluate the following limits analytically.
1.
=
2.
=
3.
lim
x2 − 5x + 4
x→ 0 x2 − 2x − 8
(𝟎)𝟐 − 𝟓(𝟎) + 𝟒
𝟒
𝟏
=
=
−
(𝟎)𝟐 − 𝟐(𝟎) − 𝟖 −𝟖
𝟐
x+1−2
=
x−3
lim
x→ 0
𝟏
𝟑
√𝟎 + 𝟏 − 𝟐 𝟏 − 𝟐 𝟏
=
=
𝟎−𝟑
−𝟑
𝟑
lim x sec x
x→ 
= 𝝅 𝒔𝒆𝒄 𝝅 = 𝝅
4.
𝟏
= −𝟐
lim
x→ 
4
= −𝝅
𝟏
𝝅
𝝅
=
=
= −𝝅
𝒄𝒐𝒔 𝝅 𝒄𝒐𝒔 𝝅 −𝟏
1 − tan x
= −√𝟐
sin x − cos x
𝒔𝒊𝒏 𝒙
𝟏 − 𝒄𝒐𝒔 𝒙
𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙
𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙
𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙
𝒄𝒐𝒔 𝒙
=
=
=
=−
𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙 𝐜𝐨 𝐬 𝒙 (𝐬𝐢 𝐧 𝒙 − 𝐜𝐨 𝐬 𝒙)
𝐜𝐨 𝐬 𝒙 (𝐜𝐨𝐬 𝐱 − 𝐬𝐢𝐧 𝐱)
=−
𝟏
𝟏
=−
𝝅 = −√𝟐
𝒄𝒐𝒔 𝒙
𝐜𝐨 𝐬 ( 𝟒 )
5.
lim g( f ( x )) ,
x→ 4
where
f ( x ) = 2 x 2 − 3 x + 1 and g( x ) = 3 x + 6 = 3
𝐥𝐢𝐦 𝟐𝒙𝟐 + 𝟑𝒙 + 𝟏 = 2(4)2 − 3(4) + 1 = 32 − 12 + 1; 𝒇(𝒙) = 𝟐𝟏
𝒙→𝟒
𝟑
𝟑
𝟑
𝐥𝐢𝐦 √𝒙 + 𝟔 = √𝟐𝟏 + 𝟔 = √𝟐𝟕 = 𝟑
𝒙→𝟒