Limits at Infinity, Part 2
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Transcript of Limits at Infinity, Part 2
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent.
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1.
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
xx + 1
xxx
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
xx
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
�x�x
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
�x�x
= limx→∞
(1 +
1
x
)
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
x
Let’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
�x�x
= limx→∞
(1 +
1
x
)This one is easy:
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
xLet’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
�x�x
= limx→∞
(1 +
1
x
)This one is easy:
limx→∞
1 +����0
1
x
A Problem That Can Be Solved In Two WaysFirst Our Basic Technique
limx→∞
x + 1
xLet’s try first our technique.First, identify the greatest exponent. It is 1. Now, divideeverything by x :
limx→∞
�x�x
+ 1x
�x�x
= limx→∞
(1 +
1
x
)This one is easy:
limx→∞
1 +����0
1
x= 1
A Problem That Can Be Solved In Two WaysNow a Faster Method
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
In this case we can directly separate the terms:
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
In this case we can directly separate the terms:
limx→∞
(x
x+
1
x
)
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
In this case we can directly separate the terms:
limx→∞
(�x
�x+
1
x
)
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
In this case we can directly separate the terms:
limx→∞
(�x
�x+
1
x
)
= limx→∞
(1 +
1
x
)
A Problem That Can Be Solved In Two WaysNow a Faster Method
Let’s try a different approach:
limx→∞
x + 1
x
In this case we can directly separate the terms:
limx→∞
(�x
�x+
1
x
)
= limx→∞
(1 +
1
x
)= 1
A Different Problem
A Different Problem
limn→∞
1 + 2 + 3 + ... + n
n2
A Different Problem
limn→∞
1 + 2 + 3 + ... + n
n2
As n→∞, the numerator becomes an infinite sum.
A Different Problem
limn→∞
1 + 2 + 3 + ... + n
n2
As n→∞, the numerator becomes an infinite sum.The numerator is an arithmetic progresion.
A Different Problem
A Different Problem
We can find a formula for this sum:
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn =
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3) + ...
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) + (n − 1 + 2) + (n − 2 + 3) + ... + (n + 1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) + (n − 2 + 3) + ... + (n + 1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) +���
���:(n + 1)
(n − 2 + 3) + ... + (n + 1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn + Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) +���
���:(n + 1)
(n − 2 + 3) + ... + (n + 1)︸ ︷︷ ︸n terms
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) +���
���:(n + 1)
(n − 2 + 3) + ... + (n + 1)︸ ︷︷ ︸n terms
= n(n+1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) +���
���:(n + 1)
(n − 2 + 3) + ... + (n + 1)︸ ︷︷ ︸n terms
= n(n+1)
2Sn = n(n + 1)
A Different Problem
We can find a formula for this sum:
Sn = 1 + 2 + ... + n
Sn = n + (n − 1) + ... + 1
Sn+Sn = (n + 1) +������:
(n + 1)(n − 1 + 2) +���
���:(n + 1)
(n − 2 + 3) + ... + (n + 1)︸ ︷︷ ︸n terms
= n(n+1)
2Sn = n(n + 1)
Sn =n(n + 1)
2
A Different Problem
A Different Problem
Now we can apply that formula to our problem:
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
n2
n2+ n
n2
2n2
n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ n
n2
2n2
n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2n2
n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2��n2
��n2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2��n2
��n2
= limn→∞
1 + 1n
2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2��n2
��n2
= limn→∞
1 +���0
1n
2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2��n2
��n2
= limn→∞
1 +���0
1n
2=
1
2
A Different Problem
Now we can apply that formula to our problem:
limn→∞
1 + 2 + 3 + ... + n
n2
limn→∞
n(n+1)2
n2= lim
n→∞
n(n + 1)
2n2
limn→∞
n2 + n
2n2
Now let’s apply our technique.Let’s divide everything by n2:
limn→∞
��n2
��n2+ �n
n�2
2��n2
��n2
= limn→∞
1 +���0
1n
2=
1
2