Lesson Explainer: Euler’s Number (𝑒) as a Limit | Nagwa Lesson Explainer: Euler’s Number (𝑒) as a Limit | Nagwa

Lesson Explainer: Euler’s Number (𝑒) as a Limit Mathematics • Third Year of Secondary School

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In this explainer, we will learn how to use the definition of 𝑒 (Euler’s number) to evaluate some special limits.

Euler’s number (𝑒=2.71828) 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 Euler’s number by using limits, and this is what we will explore in this explainer.

We can define Euler’s number using the following limit: 𝑒=1+1𝑥.lim

Using a table of values, we can see how this limit approaches Euler’s number as 𝑥 increases.

𝑥1+1𝑥
11+11=2
101+110=2.59374
1001+1100=2.70481
1‎ ‎0001+11000=2.71692
10‎ ‎0001+110000=2.71814
100‎ ‎0001+1100000=2.71826

We can use this limit to help evaluate limits and solve problems involving limits of this form. Before we look at any examples, let us consider another limit that also results in Euler’s number.

We will try substituting 𝑥=1𝑦 into our previous limit. Since we are considering the limit as 𝑥 and 𝑥=1𝑦, we can say that as 𝑥, 𝑦0. Substituting these into 𝑒=1+1𝑥lim, we will obtain 𝑒=(1+𝑦).lim

We now know two limits that evaluate to give Euler’s number. Let us summarize the results we have just shown.

Definition: Euler’s Number as a Limit

𝑒=(1+𝑥)𝑒=1+1𝑛limandlim

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 Euler’s Constant

Determine lim1+1𝑥.

Answer

We could first try to evaluate this limit directly. Our limit has 𝑥 approaching infinity, which means that the denominator of 1𝑥 is growing without bound, while the numerator remains constant, so 1𝑥 is approaching 0. This means the expression inside our parentheses is approaching 1. However, our exponent (4𝑥) is approaching infinity as 𝑥 approaches infinity.

So, we get lim1+1𝑥=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 Euler’s number: 𝑒=1+1𝑥.lim

The difference between the two limit expressions is that the one we were given has an exponent of 4𝑥 rather than 𝑥. We can use the laws of exponents to re-express it as follows: limlim1+1𝑥=1+1𝑥.

Before we can substitute Euler’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: limlim1+1𝑥=1+1𝑥.

The limit inside our exponent exists because it is just our limit result for Euler’s number 𝑒. So, we use our limit result and replace the limit inside the parentheses with 𝑒, giving us limlim1+1𝑥=1+1𝑥=𝑒.

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(𝑥+1).

Answer

Since we are asked to evaluate a limit, we could begin by attempting to do this directly. As 𝑥 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 1, so we will need to try another method.

We can see that our limit is similar to one of our limit results involving Euler’s number, which is lim(1+𝑥)=𝑒.

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

To do this, we want our exponent 1110𝑥 to be the same as that of the limit result, which is 1𝑥. To do this, we will start by using our laws of exponents to rewrite our limit: limlim(𝑥+1)=(1+𝑥), where we reorder the terms inside our parentheses and use the fact that 1110𝑥=1𝑥1110.

At this point, we want to use our limit result involving Euler’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(1+𝑥)=(1+𝑥), and we know that this is true because the limit inside our parentheses is exactly the same as the limit result involving Euler’s number. Substituting this limit being equal to 𝑒, we get (1+𝑥)=𝑒.lim

It is not always possible to directly use our limit results for Euler’s number 𝑒. 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 𝑒, 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 lim17𝑥.

Answer

We are asked to evaluate a limit that we could attempt to evaluate directly. Thus, as 𝑥 approaches , the expression inside our parentheses approaches 1 and the exponent is growing without bound. Hence, we have lim17𝑥=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 Euler’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 𝑒=(1+𝑛).lim

However, it is also possible to use 𝑒=1+1𝑛.lim

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 1𝑛.

To write our limit in this form, we will want to use a substitution. We would like 1+𝑛 inside our parentheses, so we use the substitution 𝑛=7𝑥.

We can rearrange this substitution to find 𝑥 in terms of 𝑛 so that 𝑥=7𝑛.

Multiplying this through by 5, we then have 5𝑥=35𝑛.

By using this substitution, we can rewrite our limit as limlim17𝑥=(1+𝑛).

However, this is the limit as 𝑥 approaches infinity and we want to know what happens in terms of 𝑛, so we will need to look at our substitution. As 𝑥 approaches infinity, 7𝑥 approaches 0, and since 𝑛=7𝑥, we must also have that 𝑛 approaches 0.

This gives us limlim(1+𝑛)=(1+𝑛).

Now, we use one of the laws of exponents: limlim(1+𝑛)=(1+𝑛).

Finally, we apply the power rule for limits: limlim(1+𝑛)=(1+𝑛), which we know we can do in this case because the resulting limit is our limit result involving Euler’s number.

All we need to do now is substitute this limit for 𝑒 and rearrange, so that finally we have (1+𝑛)=𝑒=1𝑒.lim

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 lim𝑥+4𝑥4.

Answer

We could try evaluating this limit directly. Inside our parentheses, we have a rational function, and we know that as 𝑥 approaches , 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 lim𝑥+4𝑥4=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 Euler’s number: 𝑒=1+1𝑛.lim

We will begin by rewriting our rational function: limlimlim𝑥+4𝑥4=𝑥4+8𝑥4=1+8𝑥4.

If we compare the two limits, we see we will need to use a substitution. We want 1+1𝑛 inside our parentheses, so we use the substitution 1𝑛=8𝑥4.

We see when 𝑥 approaches infinity that 8𝑥4 is approaching 0, so 𝑛 must also approach infinity.

Before we use this substitution, we also need to rearrange to find 𝑥 in terms of 𝑛, which we can do as follows.

We take the reciprocal of both sides of our substitution, giving us 𝑛=𝑥48.

Then, we multiply through by 8 and add 4 to both sides: 𝑥=8𝑛+4.

We can now use this substitution to rewrite our limit: limlimlim1+8𝑥4=1+1𝑛=1+1𝑛.

We want to use our limit result, but we first need our exponent to be 𝑛. To do this, we are first going to need to use the laws of exponents combined with the product rule for limits so that limlimlim1+1𝑛=1+1𝑛1+1𝑛.

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: lim1+1𝑛=1.

Next, to evaluate our other limit, we use the laws of exponents and the power rule for limits, limlim1+1𝑛=1+1𝑛, 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 Euler’s constant 𝑒: 1+1𝑛=𝑒.lim

Hence, we have shown lim𝑥+4𝑥4=𝑒.

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 limtan4𝑥+1cot.

Answer

We could try evaluating this limit directly. Inside our parentheses, we have a continuous function, so we can just substitute 𝑥=0. However, our exponent is growing without bound, so we have limtanlimtan4𝑥+1=40+1=1,cot 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 Euler’s number, that is, 𝑒=(1+𝑛).lim

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+𝑛. To do this, we begin by substituting 𝑛=4𝑥.tan

We know that as 𝑥 approaches 0, 4𝑥tan will approach 0 by direct substitution, so 𝑛 must also approach 0. Also, by taking the reciprocal of both sides of our substitution and rearranging, we have 4𝑛=1𝑥=𝑥.tancot

Using all of this, we can rewrite our limit as limtanlim4𝑥+1=(1+𝑛).cot

We can then evaluate this limit by using the laws of exponents and the power rule for limits to obtain the required exponent of 1𝑛: limlimlim(1+𝑛)=(1+𝑛)=(1+𝑛).

Of course, this is provided that the limit inside our parentheses exists, which as we know it does, since lim(1+𝑛)=𝑒.

Finally, we can use our limit result to evaluate the limit inside our parentheses as Euler’s constant: (1+𝑛)=𝑒=1𝑒.lim

Hence, we were able to show limtan4𝑥+1=1𝑒.cot

So far, we have explored the limits that result in Euler’s number. We will now consider some limits that result in the inverse of the function 𝑓(𝑥)=𝑒. When we consider any exponential function, 𝑝(𝑥)=𝑎, we know that its inverse is a logarithmic function with base 𝑎, 𝑞(𝑥)=𝑥log. So when we consider the inverse of the function 𝑓(𝑥)=𝑒, we know that it will be the logarithmic function with a base 𝑒, 𝑔(𝑥)=𝑥log, the natural logarithm. The natural logarithm function can be written as 𝑔(𝑥)=(𝑥)ln. Here is a graph showing the exponential function and the natural logarithm function. We can see how they are a reflection of each other in the line 𝑦=𝑥.

Before we define the natural logarithm as a limit, let us first recall some useful properties:

  1. 𝑦=𝑥ln is equivalent to 𝑒=𝑥,
  2. 𝑒=𝑥ln,
  3. ln𝑒=1,
  4. ln1=0,
  5. loglnln𝑥=𝑥𝑎.
    For each 𝑥,𝑦 and 𝑛,
  6. lnlnln𝑥𝑦=𝑥+𝑦,
  7. lnlnln𝑥𝑦=𝑥𝑦,
  8. lnln𝑥=𝑛𝑥,
  9. lnlog𝑥×𝑒=1.

We will begin by considering the equation

𝑦=𝑎1,(1)

where 𝑥 and 𝑦 are variables and 𝑎 is a real-valued constant. For this equation, we can see that when 𝑥0, 𝑎1 and so 𝑦0 too. We can move the 1 to the other side of the equation and take the natural logarithm of both sides of this equation to obtain lnln𝑎=(𝑦+1).

Using the properties of logarithms, we can rearrange this to

𝑥𝑎=(𝑦+1)𝑥=(𝑦+1)𝑎.lnlnlnln(2)

Let us now consider the following limit: lim𝑎1𝑥.

Using (1) and (2), we can rewrite this as limlimlnln𝑎1𝑥=𝑦𝑎(𝑦+1).

Note that, as mentioned earlier, when 𝑥0, 𝑦0 also. We can rearrange the right-hand side of this equation using properties of logarithms and limits as follows: limlimlnlnlimlnlnlimlnlnlnlimlnlnlnlim𝑎1𝑥=𝑦𝑎(𝑦+1)=𝑎(𝑦+1)=𝑎(𝑦+1)=𝑎(𝑦+1)=𝑎(𝑦+1).

Now, we may notice that the limit in the denominator of the fraction is the limit that is equal to Euler’s number. Hence, we can say that limlnlnln𝑎1𝑥=𝑎𝑒=𝑎.

We can use this limit definition of a natural logarithm to help us solve problems. Another couple of limit definitions that may also help us solve problems are limloglogandlimln(𝑥+1)𝑥=𝑒(𝑥+1)𝑥=1.

We can summarize the limits resulting in logarithms below.

Definition: Logarithms as Limits

limlnlimlogloglimln𝑎1𝑥=𝑎,(𝑥+1)𝑥=𝑒,(𝑥+1)𝑥=1.

Let us now look at some examples of how we can use these limits to solve problems.

Example 6: Evaluating a Limit by Using the Natural Logarithm

Determine lim712𝑥.

Answer

We have been given a limit to evaluate here, so we can first attempt to do this directly. If we substitute 0 into our limit, we will obtain 00, which is not defined. Hence, we will need to use some other method to determine this limit.

When we look at this limit, we may notice that it looks very similar to a limit that evaluated to the natural logarithm. Let us compare it to this limit: limln𝑎1𝑥=𝑎.

We notice that this is very similar to the limit we have been asked to evaluate. The constant 𝑎 is equal to 7. We also have slight differences in the exponent of the power and denominator of the fraction. The power is 3𝑥 instead of 𝑥 and the denominator is 2𝑥 instead of 𝑥. In order to evaluate the limit, we need to try and manipulate it to be more similar to the limit we have been given.

Let us start by changing the exponent. We can do this by using a substitution. We want to make it so that the exponent is just a variable and not a variable multiplied by a constant. We can use the substitution 3𝑥=𝑢, which is equivalent to 𝑥=𝑢3.

Before we do this substitution, we need to consider what will happen to the limit. In the limit given in the question, we are considering when 𝑥0. We are using the substitution 𝑥=𝑢3, so we can see that when 𝑥 tends to 0, so will 𝑢. Hence, when 𝑥0, 𝑢0 also. We can now carry out our substitution as follows: limlim712𝑥=71.

We can now see that the exponent of the power is just 𝑢. We have removed the coefficient using substitution. This is now looking very similar to the form we need in order to evaluate it. We just need to deal with the coefficient of 𝑢 in the denominator of the fraction. We can start by moving the constant in front of the fraction: limlim71=3271𝑢.

Now, using the limit laws, we are able to move a constant coefficient out in front of the limit: limlim3271𝑢=3271𝑢.

Now we can see that our limit is in the form of limln𝑎1𝑥=𝑎, so we can evaluate it to obtain our solution: limlimln712𝑥=3271𝑢=327.

We will now look at one final example in this explainer.

Example 7: Evaluating a Limit by Using the Natural Logarithm

Determine limln(𝑥1)𝑥2.

Answer

We could try and solve this question using direct substitution. However, if we try substituting 𝑥=2 into the limit, we will obtain 00, which is undefined. We will need to use another method in order to evaluate this limit. We notice that it looks similar to a limit that we know how to evaluate, namely, limln(𝑥+1)𝑥=1.

We notice that there are a few differences between these two limits. These are the value at which the limit is being evaluated and with some constant terms being added in both the numerator and the denominator. We can perform a substitution in our original limit to try and get it to look like the limit we know how to evaluate. We will substitute 𝑥2=𝑢, which is equivalent to 𝑥=𝑢+2.

Before we perform this substitution, we need to consider what will happen to the limiting value of 𝑥2. Since we have that 𝑢=𝑥2, we can say that as 𝑥2, 𝑢0. We are now ready to substitute: limlnlimln(𝑥1)𝑥2=(𝑢+1)𝑢.

After this substitution, we can see that the limit is exactly of the form limln(𝑥+1)𝑥=1. Hence, we can say that limln(𝑥1)𝑥2=1.

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

Key Points

  • We found and proved two useful limit results involving Euler’s number: 𝑒=(1+𝑥)𝑒=1+1𝑛.limandlim
  • We found and proved three useful limit results involving the natural logarithm: limlnlimlogloglimln𝑎1𝑥=𝑎,(𝑥+1)𝑥=𝑒,(𝑥+1)𝑥=1.
  • 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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