Lesson Explainer: Eulerโ€™s Number (e) as a Limit Mathematics

In this explainer, we will learn how to use the definition of ๐‘’ (Eulerโ€™s number) to evaluate some special limits.

Eulerโ€™s number (๐‘’=2.71828โ€ฆ) is very useful, and arises in many different branches of mathematics including the calculation of compound interest, optimization problems, calculus, and in the definition of the function representing the standard normal probability distribution.

The number may have originally been found when searching for an exponential function that differentiates into itself. However, we can also find Eulerโ€™s number by using limits, and this is what we will explore in this explainer.

To define Eulerโ€™s number as a limit, let us first recall some information about this number:

  1. lnlog(๐‘ฅ)=(๐‘ฅ)๏Œพ and ln(๐‘ฅ) is the inverse function of ๐‘’๏—.
  2. If ๐‘“(๐‘ฅ)=(๐‘ฅ)ln, then ๐‘“โ€ฒ(๐‘ฅ)=1๐‘ฅ.
  3. If ๐‘“(๐‘ฅ)=(๐‘ฅ)ln, then ๐‘“โ€ฒ(1)=1.
  4. ln(๐‘ฅ) is continuous across its entire domain.

We will use the second result to generate our limit result by using the definition of a derivative from first principles. The third result is found by substituting ๐‘ฅ=1 into the second result.

The definition of a derivative from first principles on ๐‘“(๐‘ฅ) at ๐‘ฅ=1 tells us ๐‘“โ€ฒ(1)=๏€ฝ๐‘“(1+โ„Ž)โˆ’๐‘“(1)โ„Ž๏‰.lim๏‚โ†’๏Šฆ

Replacing โ„Ž in this limit with ๐‘ฅ and using the fact that ๐‘“(๐‘ฅ)=(๐‘ฅ)ln, we have ๐‘“โ€ฒ(1)=๏€ฝ(1+๐‘ฅ)โˆ’(1)๐‘ฅ๏‰.limlnln๏—โ†’๏Šฆ

Since ln(1)=0, this simplifies to ๐‘“โ€ฒ(1)=๏€ฝ(1+๐‘ฅ)๐‘ฅ๏‰.limln๏—โ†’๏Šฆ

Using the power rule for logarithms, we can express this limit as limlnlimlnlimln๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏€ฝ(1+๐‘ฅ)๐‘ฅ๏‰=๏€ผ1๐‘ฅโ‹…(1+๐‘ฅ)๏ˆ=๏€ฝ๏€ฝ[1+๐‘ฅ]๏‰๏‰.๏Ž ๏‘

Remember, we know this limit is equal to ๐‘“โ€ฒ(1) and we already know that ๐‘“โ€ฒ(1)=1. This means we have shown that 1=๏€ฝ๏€ฝ[1+๐‘ฅ]๏‰๏‰.limln๏—โ†’๏Šฆ๏Ž ๏‘

Inside this limit, we are taking the natural logarithm, which is a continuous function. We also know that the limit is convergent, and we can use these two facts to take the natural logarithm function outside of the limit, giving us 1=๏€ฝ(1+๐‘ฅ)๏‰.lnlim๏—โ†’๏Šฆ๏Ž ๏‘

We can then simplify this further by writing both sides of this equation as exponents of ๐‘’: ๐‘’=(1+๐‘ฅ),lim๏—โ†’๏Šฆ๏Ž ๏‘ where we use the fact that the exponential function and natural logarithm functions are inverses and ๐‘’=๐‘’๏Šง.

This is our first limit result and we can immediately see its use. If we were to try and evaluate the limit directly, we would get the following indeterminate forms: limandlim๏—โ†’๏Šฆโˆž๏—โ†’๏Šฆ๏Šฑโˆž๏Žฉ๏Ž ๏‘๏Žช๏Ž ๏‘(1+๐‘ฅ)=1(1+๐‘ฅ)=1.

Limits that cannot be evaluated directly, but can be expressed in this form, can be evaluated in terms of Eulerโ€™s number, ๐‘’.

Before we move on, there is one more limit result that we can show directly from the previous one. Substituting ๐‘ฅ=1๐‘› into our limit result above gives us ๐‘’=(1+๐‘ฅ)=๏€ผ1+1๐‘›๏ˆ.limlim๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏Š๏Ž ๏‘

However, we now have a limit involving both ๐‘ฅ and ๐‘› and we want this written entirely in terms of ๐‘›. Since ๐‘ฅ=1๐‘›, when ๐‘ฅ approaches 0, 1๐‘› must also approach 0. This could be the case if ๐‘› approached either positive or negative infinity, or even oscillated between the two, which means we cannot state which specific value ๐‘› is approaching in our limit.

We can solve this problem by recalling that lim๏—โ†’๏Šฆ(1+๐‘ฅ)๏Ž ๏‘ is convergent, so both its left and right limits at 0 must be equal to the value of this limit, that is, ๐‘’. We will use the right-hand limit: ๐‘’=(1+๐‘ฅ)=(1+๐‘ฅ).limlim๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏Ž ๏‘๏Žฉ๏Ž ๏‘

Now, as ๐‘ฅ approaches 0 from the right, 1๐‘› will be approaching positive infinity.

This gives us our second limit result: ๐‘’=๏€ผ1+1๐‘›๏ˆ.lim๏Šโ†’โˆž๏Š

We can summarize the results we have just shown.

Definition: Eulerโ€™s Number as a Limit

๐‘’=(1+๐‘ฅ)๐‘’=๏€ผ1+1๐‘›๏ˆlimandlim๏—โ†’๏Šฆ๏Šโ†’โˆž๏Š๏Ž ๏‘

Let us see some examples of how we can use these two results to evaluate limits that we could not before.

Example 1: Evaluating a Limit Using Eulerโ€™s Constant

Determine lim๏—โ†’โˆž๏Šช๏—๏€ผ1+1๐‘ฅ๏ˆ.

Answer

We could first try to evaluate this limit directly. Our limit has ๐‘ฅ approaching infinity, which means that the denominator of 1๐‘ฅ is growing without bound, while the numerator remains constant, so 1๐‘ฅ is approaching 0. This means the expression inside our parentheses is approaching 1. However, our exponent (4๐‘ฅ) is approaching infinity as ๐‘ฅ approaches infinity.

So, we get lim๏—โ†’โˆž๏Šช๏—โˆž๏€ผ1+1๐‘ฅ๏ˆ=1, which is an indeterminate form. This means we are going to need to try some other method to evaluate our limit.

We notice that the limit we have been given is very similar to one that expresses the value of Eulerโ€™s number: ๐‘’=๏€ผ1+1๐‘ฅ๏ˆ.lim๏—โ†’โˆž๏—

The difference between the two limit expressions is that the one we were given has an exponent of 4๐‘ฅ rather than ๐‘ฅ. We can use the laws of exponents to re-express it as follows: limlim๏—โ†’โˆž๏Šช๏—๏—โ†’โˆž๏—๏Šช๏€ผ1+1๐‘ฅ๏ˆ=๏€ฝ๏€ผ1+1๐‘ฅ๏ˆ๏‰.

Before we can substitute Eulerโ€™s number into the limit expression, we need to move the exponent of 4 outside the limit. Provided the new limit exists, we can use the power rule for limits to achieve this: limlim๏—โ†’โˆž๏—๏Šช๏—โ†’โˆž๏—๏Šช๏€ฝ๏€ผ1+1๐‘ฅ๏ˆ๏‰=๏€ฝ๏€ผ1+1๐‘ฅ๏ˆ๏‰.

The limit inside our exponent exists because it is just our limit result for Eulerโ€™s number ๐‘’. So, we use our limit result and replace the limit inside the parentheses with ๐‘’, giving us limlim๏—โ†’โˆž๏Šช๏—๏—โ†’โˆž๏—๏Šช๏Šช๏€ผ1+1๐‘ฅ๏ˆ=๏€ฝ๏€ผ1+1๐‘ฅ๏ˆ๏‰=๐‘’.

Our next example shows how we can use our other limit result to help us evaluate a limit.

Example 2: Solving Limits by Transforming Them into the Natural Exponent Limit Forms

Determine lim๏—โ†’๏Šฆ(๐‘ฅ+1)๏Ž ๏Ž ๏Ž ๏ŽŸ๏‘.

Answer

Since we are asked to evaluate a limit, we could begin by attempting to do this directly. As ๐‘ฅ approaches 0, the expression inside our parentheses approaches 1 and the magnitude of our exponent is growing without bound. This is an indeterminate form, specifically 1โˆž, so we will need to try another method.

We can see that our limit is similar to one of our limit results involving Eulerโ€™s number, which is lim๏—โ†’๏Šฆ(1+๐‘ฅ)=๐‘’.๏Ž ๏‘

So, we can try using this result to help us evaluate our limit.

To do this, we want our exponent 1110๐‘ฅ to be the same as that of the limit result, which is 1๐‘ฅ. To do this, we will start by using our laws of exponents to rewrite our limit: limlim๏—โ†’๏Šฆ๏—โ†’๏Šฆ(๐‘ฅ+1)=๏€ฝ(1+๐‘ฅ)๏‰,๏Ž ๏Ž ๏Ž ๏ŽŸ๏‘๏Ž ๏‘๏Ž ๏Ž ๏Ž ๏ŽŸ where we reorder the terms inside our parentheses and use the fact that 1110๐‘ฅ=1๐‘ฅโ‹…1110.

At this point, we want to use our limit result involving Eulerโ€™s number; however, we first need to take our exponent outside of our limit altogether and to do that we need to use the power rule for limits.

This tells us we can take the exponent outside of the limit provided our new limit exists.

In our case, we have limlim๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏€ฝ(1+๐‘ฅ)๏‰=๏€ฝ(1+๐‘ฅ)๏‰,๏Ž ๏‘๏Ž ๏Ž ๏Ž ๏ŽŸ๏Ž ๏‘๏Ž ๏Ž ๏Ž ๏ŽŸ and we know that this is true because the limit inside our parentheses is exactly the same as the limit result involving Eulerโ€™s number. Substituting this limit being equal to ๐‘’, we get ๏€ฝ(1+๐‘ฅ)๏‰=๐‘’.lim๏—โ†’๏Šฆ๏Ž ๏‘๏Ž ๏Ž ๏Ž ๏ŽŸ๏Ž ๏Ž ๏Ž ๏ŽŸ

It is not always possible to directly use our limit results for Eulerโ€™s number ๐‘’. We may have to use other tools such as polynomial division, factoring, or substitution. However, the basic premise is the same, we take a limit we are unable to evaluate and write it in a form for ๐‘’, which we can use then our limit results to evaluate.

Example 3: Evaluating a Limit by Transforming It into the Natural Exponent Limit Form

Determine lim๏—โ†’โˆž๏Šซ๏—๏€ผ1โˆ’7๐‘ฅ๏ˆ.

Answer

We are asked to evaluate a limit that we could attempt to evaluate directly. Thus, as ๐‘ฅ approaches โˆž, the expression inside our parentheses approaches 1 and the exponent is growing without bound. Hence, we have lim๏—โ†’โˆž๏Šซ๏—โˆž๏€ผ1โˆ’7๐‘ฅ๏ˆ=1.

This is an indeterminate form, so we will need to try another method to evaluate this limit.

This limit is similar to one of the limit results involving Eulerโ€™s number, so we can try using this result to help us evaluate our limit. We have a lot of choices in how to go about this.

We are going to try and write this limit in a form where we can use ๐‘’=(1+๐‘›).lim๏Šโ†’๏Šฆ๏Ž ๏‘ƒ

However, it is also possible to use ๐‘’=๏€ผ1+1๐‘›๏ˆ.lim๏Šโ†’โˆž๏Š

Usually, one of the limit results ends up being easier than the other and it can be very difficult to tell which limit result to use just by looking at the question, so if we get stuck using one result, we can always try using the other limit that is in the form with exponent 1๐‘›.

To write our limit in this form, we will want to use a substitution. We would like 1+๐‘› inside our parentheses, so we use the substitution ๐‘›=โˆ’7๐‘ฅ.

We can rearrange this substitution to find ๐‘ฅ in terms of ๐‘› so that ๐‘ฅ=โˆ’7๐‘›.

Multiplying this through by 5, we then have 5๐‘ฅ=โˆ’35๐‘›.

By using this substitution, we can rewrite our limit as limlim๏—โ†’โˆž๏Šซ๏—๏—โ†’โˆž๏€ผ1โˆ’7๐‘ฅ๏ˆ=(1+๐‘›).๏Žช๏Žข๏Žค๏‘ƒ

However, this is the limit as ๐‘ฅ approaches infinity and we want to know what happens in terms of ๐‘›, so we will need to look at our substitution. As ๐‘ฅ approaches infinity, โˆ’7๐‘ฅ approaches 0, and since ๐‘›=โˆ’7๐‘ฅ, we must also have that ๐‘› approaches 0.

This gives us limlim๏—โ†’โˆž๏Šโ†’๏Šฆ(1+๐‘›)=(1+๐‘›).๏Žช๏Žข๏Žค๏‘ƒ๏Žช๏Žข๏Žค๏‘ƒ

Now, we use one of the laws of exponents: limlim๏Šโ†’๏Šฆ๏Šโ†’๏Šฆ๏Šฑ๏Šฉ๏Šซ(1+๐‘›)=๏€ฝ(1+๐‘›)๏‰.๏Žช๏Žข๏Žค๏‘ƒ๏Ž ๏‘ƒ

Finally, we apply the power rule for limits: limlim๏Šโ†’๏Šฆ๏Šฑ๏Šฉ๏Šซ๏Šโ†’๏Šฆ๏Šฑ๏Šฉ๏Šซ๏€ฝ(1+๐‘›)๏‰=๏€ฝ(1+๐‘›)๏‰,๏Ž ๏‘ƒ๏Ž ๏‘ƒ which we know we can do in this case because the resulting limit is our limit result involving Eulerโ€™s number.

All we need to do now is substitute this limit for ๐‘’ and rearrange, so that finally we have ๏€ฝ(1+๐‘›)๏‰=๐‘’=1๐‘’.lim๏Šโ†’๏Šฆ๏Šฑ๏Šฉ๏Šซ๏Šฑ๏Šฉ๏Šซ๏Šฉ๏Šซ๏Ž ๏‘ƒ

In our next example, we will consider the limit of a rational function raised to a linear exponent.

Example 4: Solving Limits by Transforming Them into the Natural Exponent Limit Forms

Determine lim๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏€ผ๐‘ฅ+4๐‘ฅโˆ’4๏ˆ.

Answer

We could try evaluating this limit directly. Inside our parentheses, we have a rational function, and we know that as ๐‘ฅ approaches โˆž, we can see what happens by looking at the quotient of the leading terms in our rational function. Since this is equal to 1, the limit of our rational function is also 1. However, our exponent is growing without bound, so we have lim๏—โ†’โˆž๏—๏Šฑ๏Šฉโˆž๏€ผ๐‘ฅ+4๐‘ฅโˆ’4๏ˆ=1, which is an indeterminate form. We will therefore need to try another method to evaluate this limit.

Instead, let us try and evaluate this by using a limit result involving Eulerโ€™s number: ๐‘’=๏€ผ1+1๐‘›๏ˆ.lim๏Šโ†’โˆž๏Š

We will begin by rewriting our rational function: limlimlim๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏€ผ๐‘ฅ+4๐‘ฅโˆ’4๏ˆ=๏€ผ๐‘ฅโˆ’4+8๐‘ฅโˆ’4๏ˆ=๏€ผ1+8๐‘ฅโˆ’4๏ˆ.

If we compare the two limits, we see we will need to use a substitution. We want 1+1๐‘› inside our parentheses, so we use the substitution 1๐‘›=8๐‘ฅโˆ’4.

We see when ๐‘ฅ approaches infinity that 8๐‘ฅโˆ’4 is approaching 0, so ๐‘› must also approach infinity.

Before we use this substitution, we also need to rearrange to find ๐‘ฅ in terms of ๐‘›, which we can do as follows.

We take the reciprocal of both sides of our substitution, giving us ๐‘›=๐‘ฅโˆ’48.

Then, we multiply through by 8 and add 4 to both sides: ๐‘ฅ=8๐‘›+4.

We can now use this substitution to rewrite our limit: limlimlim๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏Šโ†’โˆž๏Šฎ๏Š๏Šฐ๏Šช๏Šฑ๏Šฉ๏Šโ†’โˆž๏Šฎ๏Š๏Šฐ๏Šง๏€ผ1+8๐‘ฅโˆ’4๏ˆ=๏€ผ1+1๐‘›๏ˆ=๏€ผ1+1๐‘›๏ˆ.

We want to use our limit result, but we first need our exponent to be ๐‘›. To do this, we are first going to need to use the laws of exponents combined with the product rule for limits so that limlimlim๏Šโ†’โˆž๏Šฎ๏Š๏Šฐ๏Šง๏Šโ†’โˆž๏Šฎ๏Š๏Šโ†’โˆž๏€ผ1+1๐‘›๏ˆ=๏€ผ1+1๐‘›๏ˆโ‹…๏€ผ1+1๐‘›๏ˆ.

To use the product rule for limits, we need the limit of both factors to exist. We will show in our working that both of these limits exist.

We can evaluate one of these limits directly: lim๏Šโ†’โˆž๏€ผ1+1๐‘›๏ˆ=1.

Next, to evaluate our other limit, we use the laws of exponents and the power rule for limits, limlim๏Šโ†’โˆž๏Šฎ๏Š๏Šโ†’โˆž๏Š๏Šฎ๏€ผ1+1๐‘›๏ˆ=๏€ฝ๏€ผ1+1๐‘›๏ˆ๏‰, and this is true provided this limit exists, which we know it does because it is our previous limit result. This means we can replace this limit with Eulerโ€™s constant ๐‘’: ๏€ฝ๏€ผ1+1๐‘›๏ˆ๏‰=๐‘’.lim๏Šโ†’โˆž๏Š๏Šฎ๏Šฎ

Hence, we have shown lim๏—โ†’โˆž๏—๏Šฑ๏Šฉ๏Šฎ๏€ผ๐‘ฅ+4๐‘ฅโˆ’4๏ˆ=๐‘’.

We can also use these results to solve limits involving more complicated functions.

Example 5: Evaluating Limits by Transforming Them into the Natural Exponent Limit Forms

Determine limtan๏—โ†’๏Šฆ๏Šฉ๏—๏€นโˆ’4๐‘ฅ+1๏…cot๏Žข.

Answer

We could try evaluating this limit directly. Inside our parentheses, we have a continuous function, so we can just substitute ๐‘ฅ=0. However, our exponent is growing without bound, so we have limtanlimtan๏—โ†’๏Šฆ๏Šฉ๏—๏—โ†’๏Šฆ๏Šฉโˆžโˆž๏€นโˆ’4๐‘ฅ+1๏…=๏€นโˆ’40+1๏…=1,cot๏Žข which is an indeterminate form, so we will need to try another method to evaluate this limit.

Instead, let us try and evaluate this by using a limit result involving Eulerโ€™s number, that is, ๐‘’=(1+๐‘›).lim๏Šโ†’๏Šฆ๏Ž ๏‘ƒ

In order to compare this with the limit we are asked to find, we will need to manipulate the expression inside our parentheses into the form 1+๐‘›. To do this, we begin by substituting ๐‘›=โˆ’4๐‘ฅ.tan๏Šฉ

We know that as ๐‘ฅ approaches 0, โˆ’4๐‘ฅtan๏Šฉ will approach 0 by direct substitution, so ๐‘› must also approach 0. Also, by taking the reciprocal of both sides of our substitution and rearranging, we have โˆ’4๐‘›=1๐‘ฅ=๐‘ฅ.tancot๏Šฉ๏Šฉ

Using all of this, we can rewrite our limit as limtanlim๏—โ†’๏Šฆ๏Šฉ๏—๏Šโ†’๏Šฆ๏€นโˆ’4๐‘ฅ+1๏…=(1+๐‘›).cot๏Žข๏Žช๏Žฃ๏‘ƒ

We can then evaluate this limit by using the laws of exponents and the power rule for limits to obtain the required exponent of 1๐‘›: limlimlim๏Šโ†’๏Šฆ๏Šโ†’๏Šฆ๏Šฑ๏Šช๏Šโ†’๏Šฆ๏Šฑ๏Šช(1+๐‘›)=๏€ฝ(1+๐‘›)๏‰=๏€ฝ(1+๐‘›)๏‰.๏Žช๏Žฃ๏‘ƒ๏Ž ๏‘ƒ๏Ž ๏‘ƒ

Of course, this is provided that the limit inside our parentheses exists, which as we know it does, since lim๏Šโ†’๏Šฆ(1+๐‘›)=๐‘’.๏Ž ๏‘ƒ

Finally, we can use our limit result to evaluate the limit inside our parentheses as Eulerโ€™s constant: ๏€ฝ(1+๐‘›)๏‰=๐‘’=1๐‘’.lim๏Šโ†’๏Šฆ๏Šฑ๏Šช๏Šฑ๏Šช๏Šช๏Ž ๏‘ƒ

Hence, we were able to show limtan๏—โ†’๏Šฆ๏Šฉ๏—๏Šช๏€นโˆ’4๐‘ฅ+1๏…=1๐‘’.cot๏Žข

Let us finish by recapping some of the basic points of this explainer.

Key Points

  • We found and proved two useful limit results involving Eulerโ€™s number: ๐‘’=(1+๐‘ฅ)๐‘’=๏€ผ1+1๐‘›๏ˆ.limandlim๏—โ†’๏Šฆ๏Šโ†’โˆž๏Š๏Ž ๏‘
  • We may use these results to evaluate limits that give indeterminate forms by direct substitution or evaluation.
  • In order to use these results, we may sometimes need to manipulate our limit using techniques such as polynomial division, substitution, or factoring.

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