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In this presentation I present the concept of limits at infinity and the basic technique for solving them. We solve an example step by step.
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What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
What are Limits at Infinity?
Limits at infinity are just like normal limits.
The difference is that x is approaching infinity instead of a number.
limx→∞
f (x)
This means that x is growing without bounds.
Or that it takes values greater than any number you can come upwith.
The Basic Technique For Solving Limits at Infinity
1. First of all, we find the greatest exponent in our function.
I For example:
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
I For example:
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.
I For example:
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.Here n is the greatest exponent we found before.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.Here n is the greatest exponent we found before.
I In our example:
The Basic Technique For Solving Limits at Infinity
To solve these limits we use a basic technique:
1. First of all, we find the greatest exponent in our function.I For example:
2x2 − 1
x2 + x
The greatest exponent is 2.
2. Then we divide both the numerator and denominator by xn.Here n is the greatest exponent we found before.
I In our example:2x2−1x2
x2+xx2
3. Now we simplify our expression algebraically:
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=2x2
x2− 1
x2
x2
x2+ x
x2
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=
2��x2
��x2− 1
x2
x2
x2+ x
x2
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=
2��x2
��x2− 1
x2
��x2
��x2+ x
x2
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=
2��x2
��x2− 1
x2
��x2
��x2+ �x
x �2
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=
2��x2
��x2− 1
x2
��x2
��x2+ �x
x �2
=2− 1
x2
1 + 1x
4. Now we calculate the limit directly.
3. Now we simplify our expression algebraically:
2x2−1x2
x2+xx2
=
2��x2
��x2− 1
x2
��x2
��x2+ �x
x �2
=2− 1
x2
1 + 1x
4. Now we calculate the limit directly.
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.
Next, we divide everything by x3:
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4x3
x3− 2x2
x3+ 1
x3
5x3
x3− 3
x3
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4��x3
��x3− 2x2
x3+ 1
x3
5x3
x3− 3
x3
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4��x3
��x3− 2��x2
x �3+ 1
x3
5x3
x3− 3
x3
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4��x3
��x3− 2��x2
x �3+ 1
x3
5��x3
��x3− 3
x3
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4��x3
��x3− 2��x2
x �3+ 1
x3
5��x3
��x3− 3
x3
And we’re left with:
The First Example
Let’s find the following limit:
limx→∞
4x3 − 2x2 + 1
5x3 − 3
First of all, we find the greatest exponent there. In this case, it is 3.Next, we divide everything by x3:
limx→∞
4��x3
��x3− 2��x2
x �3+ 1
x3
5��x3
��x3− 3
x3
And we’re left with:
limx→∞
4− 2x + 1
x3
5− 3x3
The First Example
Now we can calculate the limit directly:
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
How do we do that?!
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
How do we do that?!Just observe that anything that is divided by x will go to zero.
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
How do we do that?!Just observe that anything that is divided by x will go to zero.Why?
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
How do we do that?!Just observe that anything that is divided by x will go to zero.Why?Because x is approaching ∞!
The First Example
Now we can calculate the limit directly:
limx→∞
4− 2x + 1
x3
5− 3x3
How do we do that?!Just observe that anything that is divided by x will go to zero.Why?Because x is approaching ∞!Just try with your calculator. Divide anything by a very bignumber and you’ll get very close to 0.
The First Example
The First Example
So, using that fact, we have that:
The First Example
So, using that fact, we have that:
limx→∞
4− 2x + 1
x3
5− 3x3
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x + 1
x3
5− 3x3
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5− 3x3
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5−��70
3x3
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5−��70
3x3
And we’re left with:
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5−��70
3x3
And we’re left with:
limx→∞
4
5
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5−��70
3x3
And we’re left with:
limx→∞
4
5=
4
5
The First Example
So, using that fact, we have that:
limx→∞
4−���0
2x +��70
1x3
5−��70
3x3
And we’re left with:
limx→∞
4
5=
4
5
That’s the final answer!
