Limits
(introduction)
Alex Karassev
The concept of a limit
What happens to x2 when x
approaches 2?
 Look at table of values:

x
x2
1.9
1.99
1.999
3.61
3.9601
3.996001
2.1
2.01
2.001
4.41
4.0401
4.004001
|x2 - 4|
0.39
0.0399
0.003999
0.41
0.0401
0.004001
Direct substitution
Thus we see that x2 approaches 4, i.e.
22, as x approaches 2
 So we can just substitute 2 in the
function x2
 Is it always the case?

Example: direct substitution is
impossible
Look at (x3-1)/(x-1) as x approaches 1
 We cannot sub. x=1, however, we can
substitute the values of x near 1

x
0.9
0.99
0.999
0.9999
(x3-1)/(x-1)
2.71
2.9701
2.997001
2.99970001
x
1.1
1.01
1.001
1.0001
(x3-1)/(x-1)
3.31
3.0301
3.003001
3.00030001
Guess: (x3 - 1) / (x - 1) approaches 3 as x approaches 1
Measuring the earth…
How to decide whether a planet is flat
or "round"?
 Look at circles of the same radius on
a sphere and on the plane and
compare their circumferences

Circumferences
Sphere of radius R
Euclidean plane
r
Cr
a
R
Cr
r
θ
R
Cr = 2π a
Cr = 2π r
So Cr / r = 2π
is constant!
(does not depend on circle)
a = R sinθ
and
r=Rθ
So Cr / r = (2π a) / r = (2π R sinθ) / (R θ)
= 2π (sinθ / θ)
Depends on θ and hence on the
circle!
What if θ is very small?
r
When θ is small the piece of sphere
enclosed by our circle is "almost flat", so
we should expect that
Cr
Cr / r is very close to 2π
θ
However, Cr / r = 2π (sinθ / θ)
Therefore, when θ is close to 0,
sinθ / θ is close to 1 !
θ is close to 0
Guess: sinθ / θ approaches 1 as θ approaches 0
Equivalently: sinθ ≈ θ when θ is near 0
Example: what is sin 1° ?
sinθ ≈ θ when θ is near 0
 1° = π / 180 radians, so
θ = π / 180
 We have:
sin1° = sin (π / 180) ≈ π / 180 ≈
3.14 / 180 = 0.0174444…
 Calculator gives:
sin1° ≈ 0.017452406

Table of values for sinθ / θ
θ
sinθ / θ
0.5
0.4
0.3
0.2
0.958851077
0.973545856
0.985067356
0.993346654
0.1
0.01
0.001
0.998334166
0.999983333
0.999999833
0.0001
0.999999998
Graph of y = sinθ / θ
y
1
1
0.8
0.6
0.4
0.2
θ
π
-π
0
-0.2
-0.4
-0.6
-0.8
-1
-10
-8
-6
-4
-2
0
2
4
6
8
10
Danger of guessing
cos x
 What happens to f ( x )  sin x 
10000
when x approaches 0?
x
f(x)
0.01
0.010099828
0.001
0.0011
0.0001
0.0002
0.00001
0.00011
What is our guess?
Danger of guessing

More values…
x
f(x)
0.001
0.0011
0.0001
0.0002
0.00001
0.00011
0.000001
0.000101
0.0000001
0.0001001
0.00000001
0.00010001
cos x
f ( x)  sin x 
10000
What is our guess now?
Solution

cos x
f ( x)  sin x 
10000
As x approaches 0:
 sinx approaches sin0 = 0
 cosx approaches cos0 = 1
 So sinx + cosx / 10000 approaches
0 + 1 / 10000 = 0.0001

Informal definition

We write
lim f ( x)  L
x a
if we can make the values of f(x) as
close to L as we like by taking x to be
sufficiently close to a but not equal to a

Note: the value of function f at a, i.e.
f(a), does not affect the limit!
Examples
x 1
lim
3
x 1 x  1
3
y
4
y=f(x)
x
sin x
lim
1
x 0
x
x2 , x  2
if f ( x)  
, then :
 0, x  2
2
2
lim f ( x)  lim x  2  4  f (2)  0
x 2
x 2
2
When we compute limit as x approaches 2, x is not equal to 2
Does limit always exist?
y
0, x  0
let h( x)  
1, x  0
y=h(x)
1
x
0

If x approaches 0 from the left then h(x) is constantly 0

If x approaches 0 from the right then h(x) is constantly 1

Thus when x is close to 0 h(x) can be arbitrarily close (in fact,
even equal to) 0 or 1 and we cannot choose either of this two
numbers for the value of the limit of h(x) as x approaches 0

So lim h( x ) does not exists (D.N.E.)
x 0
One-sided limits
Limit of f(x)
as x approaches a from the left
 Limit of f(x)
as x approaches a from the right

From the left

We write
lim f ( x)  L
xa
if we can make the values of f(x) as close
to L as we like by taking x to be
sufficiently close to a and less than a
y
f(a)
y = f(x)
L
x
x
a
From the right

We write
lim f ( x)  L
xa
if we can make the values of f(x) as close
to L as we like by taking x to be
sufficiently close to a and more than a
y
f(a)
y = f(x)
L
x
a
x
Theorem
lim f ( x) exists
xa
if and only if both limits
lim f ( x) and lim f ( x) exist
xa
xa
AND lim f ( x)  lim f ( x) ( lim f ( x) )
xa
xa
xa
Infinite limits
Let f(x) = 1/x2
 If x is close 0 then1/x2 is a
very large positive number
 Therefore the finite limit of
1/x2 as x approaches 0
does not exist
 However, we may say that
1/x2 approaches positive
infinity as x approaches 0

y
M
x
0
Infinite limits
lim f ( x)  
We write x  a
if we can make the values of f(x) bigger than any
number M by taking x to be sufficiently close to a
and not equal to a
(note: think of M as of large positive number)

lim f ( x)  
We write x  a
if we can make the values of f(x) smaller than
any number M by taking x to be sufficiently close
to a and not equal to a
(note: think of M as of large negative number)

Infinite limits: one-sided limits
y
1
lim  
x 0 x
x
0
1
lim  
x 0 x
Vertical asymptote

Definition
we say that a line x=a is a vertical
asymptote for the function f if at leas
one of the following is true:
lim f ( x)   lim f ( x)  
xa
xa
lim f ( x)   lim f ( x)  
xa
xa
Example
 
f ( x)  sin  
x
x
1/2
1/3
1/4
1/n
What happens when x
approaches 0?
sin(π/x)
sin(π/1/2)=sin(2 π)=0
0
0
sin(π/1/n)=sin(πn)=0
 
Is lim sin    0 ?
x 0
x
However…
x
2/5
2/9
2/13
sin(π/x)
sin(π/2/5)=sin(5/2)π=
sin(π/2 + 2 π) =
sin(π/2) = 1
1
1
2/17
2/(4n+1)
1
1
 
Is lim sin    1 ?
x 0
x
Limit does not exist!