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