In this second day we solve the most basic limits we could find, like the limit of a constant. Then we find the limit of the sum, the product and the quotient of two functions. We solve two simple examples.
Day 3 of Free Intuitive Calculus Course: Limits by FactoringPablo Antuna
Today we focus on limits by factoring. We solve limits by factoring and cancelling. This is one of the basic techniques for solving limits. We talk about the idea behind this technique and we solve some examples step by step.
In this video we learn how to solve limits that involve trigonometric functions. It is all based on using the fundamental trigonometric limit, which is proved using the squeeze theorem.
For more lessons: http://www.intuitive-calculus.com/solving-limits.html
Watch video: http://www.youtube.com/watch?v=1RqXMJWcRIA
In this second day we solve the most basic limits we could find, like the limit of a constant. Then we find the limit of the sum, the product and the quotient of two functions. We solve two simple examples.
Day 3 of Free Intuitive Calculus Course: Limits by FactoringPablo Antuna
Today we focus on limits by factoring. We solve limits by factoring and cancelling. This is one of the basic techniques for solving limits. We talk about the idea behind this technique and we solve some examples step by step.
In this video we learn how to solve limits that involve trigonometric functions. It is all based on using the fundamental trigonometric limit, which is proved using the squeeze theorem.
For more lessons: http://www.intuitive-calculus.com/solving-limits.html
Watch video: http://www.youtube.com/watch?v=1RqXMJWcRIA
Day 5 of the Intuitive Online Calculus Course: The Squeeze TheoremPablo Antuna
In this presentation we learn about the Squeeze Theorem. We first try to get the intuition behind it, why it must be true. Then we apply it to solve the Fundamental Trigonometric Limit. This limit is very important for solving other trigonometric limits.
To solve this limit we use a little bit of geometry and then apply the Squeeze Theorem.
We solve limits by rationalizing. This is the second technique you may learn after limits by factoring. We solve two examples step by step.
Watch video: http://www.youtube.com/watch?v=8CtpuojMJzA
More videos and lessons: http://www.intuitive-calculus.com/solving-limits.html
In this video we learn how to solve limits by factoring and cancelling. This is one of the most simple and powerful techniques for solving limits.
Watch video: http://www.youtube.com/watch?v=r0Qw5gZuTYE
For more videos and lessons: http://www.intuitive-calculus.com/solving-limits.html
Youth innovators - Social media for businessBrian Jackson
For the Toronto Youth Innovators conference, a presentation about evaluating social media business brands and organizing a Twitter chat.
Refer to these articles:
http://www.itbusiness.ca/news/how-to-evaluate-your-social-media-brand-beyond-the-numbers/43667
http://www.itbusiness.ca/news/what-we-learned-from-running-our-twitter-chats/43695
Day 5 of the Intuitive Online Calculus Course: The Squeeze TheoremPablo Antuna
In this presentation we learn about the Squeeze Theorem. We first try to get the intuition behind it, why it must be true. Then we apply it to solve the Fundamental Trigonometric Limit. This limit is very important for solving other trigonometric limits.
To solve this limit we use a little bit of geometry and then apply the Squeeze Theorem.
We solve limits by rationalizing. This is the second technique you may learn after limits by factoring. We solve two examples step by step.
Watch video: http://www.youtube.com/watch?v=8CtpuojMJzA
More videos and lessons: http://www.intuitive-calculus.com/solving-limits.html
In this video we learn how to solve limits by factoring and cancelling. This is one of the most simple and powerful techniques for solving limits.
Watch video: http://www.youtube.com/watch?v=r0Qw5gZuTYE
For more videos and lessons: http://www.intuitive-calculus.com/solving-limits.html
Youth innovators - Social media for businessBrian Jackson
For the Toronto Youth Innovators conference, a presentation about evaluating social media business brands and organizing a Twitter chat.
Refer to these articles:
http://www.itbusiness.ca/news/how-to-evaluate-your-social-media-brand-beyond-the-numbers/43667
http://www.itbusiness.ca/news/what-we-learned-from-running-our-twitter-chats/43695
3. What are Indeterminate Forms?
¥/¥
0 / 0 0×¥ ¥-¥ 00
,¥0
,1¥
Applying L’Hopital’s Rule:
• Check that the limit of f(x) / g(x) is an indeterminate form 0 / 0 .
• Differentiate f and g separately.
• Find the limit of f’(x) / g’(x).
4. Indeterminate Form of Type 0 / 0
Suppose that f and g are differentiable functions on an open interval containing
x = a, except possibly at x = a, and that
if lim
x®a
f '(x)
g'(x)
é
ë
ê
ù
û
úexists, or¥ if this limit is +¥ or -¥, then
lim
x®a
f (x)= 0 and lim
x®a
g(x) = 0
lim
x®a
f (x)
g(x)
= lim
x®a
f '(x)
g'(x)
5. How about some examples?
Find the following limits:
lim
x®2
x2
- 4
x -2
= lim
x®2
2x
1
= 2×2
lim
x®
p
2
1-sin x
cosx
= lim
x®
p
2
-cosx
-sin x
=
0
-1
= 0
lim
x®0
ex
-1
x3
= lim
x®0
ex
3x2
=
1
0
= +¥
lim
x®+¥
x
-
4
3
sin
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
-
4
3
x
-
7
3
-
1
x2
cos
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
4
3
x
-
1
3
cos
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
4
3x
1
3
cos
1
x
æ
è
ç
ö
ø
÷
=
4
3 +¥
( )
1
3 cos
1
+¥
æ
è
ç
ö
ø
÷
=
4
+¥
( )cos 0
( )
=
4
+¥×1
= 0
1.
2.
3.
4.
6. Indeterminate Form of Type
Suppose that f and g are differentiable functions on an open interval containing
x = a, except possibly at x = a, and that
if lim
x®a
f '(x)
g'(x)
é
ë
ê
ù
û
úexists, or¥ if this limit is +¥ or -¥, then
lim
x®a
f (x) =¥ and lim
x®a
g(x) =¥
lim
x®a
f (x)
g(x)
= lim
x®a
f '(x)
g'(x)
¥/¥
7. How about some examples?
Find the following limits:
lim
x®+¥
x
ex
= lim
x®+¥
1
ex
=
1
+¥
= 0
lim
x®0+
ln x
csc x
= lim
x®0+
1
x
-csc xcot x
= lim
x®0+
sin x
-x
tan x = lim
x®0+
-
sin x
x
× lim
x®0+
tan x =
lim
x®0+
-
cosx
1
lim
x®0+
tan x = -1
( )× 0
( ) = 0
lim
x®0+
ln sin x
( )
ln tan x
( )
= lim
x®0+
1
sin x
×cosx
1
tan x
×sec2
x
= lim
x®0+
cot x
cot x×sec2
x
= lim
x®0+
1
sec2
x
=
lim
x®0+
cos2
x = cos2
0 =1
1.
2.
3.
8. Indeterminate Form of Type 0×¥
Can sometimes be evaluated by rewriting the product as
a ratio:
lim
x®0+
xln x = lim
x®0+
ln x
1
x
= lim
x®0+
1
x
-
1
x2
= lim
x®0+
-x2
x
= lim
x®0+
-x
( )= 0
lim
x®
p
4
1- tan x
( )sec2x = lim
x®
p
4
1- tan x
( )
cos2x
= lim
x®
p
4
-sec2
x
-2sin2x
= lim
x®
p
4
sec2
x
-2sin2x
=
sec2 p
4
æ
è
ç
ö
ø
÷
2sin 2×
p
4
æ
è
ç
ö
ø
÷
=
2
2
=1
1.
2.
9. Indeterminate Form of Type ¥-¥
Can sometimes be evaluated by combining the terms and
manipulating the result to produce quotient
lim
x®0+
1
x
-
1
sin x
æ
è
ç
ö
ø
÷ = lim
x®0+
sin x - x
xsin x
æ
è
ç
ö
ø
÷ = lim
x®0+
cosx -1
sin x + xcosx
æ
è
ç
ö
ø
÷ =
lim
x®0+
-sin x
cosx +cosx - xsin x
æ
è
ç
ö
ø
÷ =
0
1+1- 0
= 0
lim
x®0
1
x
-
1
ex
-1
æ
è
ç
ö
ø
÷ = lim
x®0
ex
-1
( )- x
x ex
-1
( )
æ
è
ç
ç
ö
ø
÷
÷
= lim
x®0
ex
- x -1
x ex
-1
( )
æ
è
ç
ç
ö
ø
÷
÷
=
lim
x®0
ex
-1
( )
ex
-1
( )+ xex
= lim
x®0
ex
ex
+ex
+ xex
= lim
x®0
ex
ex
2+ x
( )
= lim
x®0
1
x +2
=
1
2
1.
2.
10. Indeterminate Form of Type
Can sometimes be evaluated by first introducing a
dependent variable
00
,¥0
,1¥
y = f (x)g(x)
And then computing the limit of lny. Since
lny = ln f (x)g(x)
é
ë ù
û= g(x)×ln f (x)
[ ]
The limit of lny will be an indeterminate form of type 0×¥
11. How about an example?
y = 1+ x
( )
1
x
ln y = ln 1+ x
( )
1
x =
1
x
ln 1+ x
( )=
ln 1+ x
( )
x
Show that lim
x®0
1+ x
( )
1
x = e
lim
x®0
ln y
( ) = lim
x®0
ln 1+ x
( )
x
= lim
x®0
1
1+ x
1
=1
lim
x®0
ln y
( )=1
x®0
limy = e1
lim
x®0
1+ x
( )
1
x = e