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

# Lesson Video: Eulerβs Number (e) as a Limit Mathematics

In this video, we will learn how to use the definition of e (Eulerβs number) to evaluate some special limits.

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### Video Transcript

Eulerβs Number As a Limit

In this video, weβre going to discuss how we can define Eulerβs number as a limit and how we can use this limit to help us evaluate other limits.

To define Eulerβs number as a limit, weβre first going to need to recall some information. The first thing we need to recall is if we have a function π of π₯ which is the natural logarithm of π₯, then weβve shown π prime of π₯ is the reciprocal function one over π₯. And weβve seen that this is true from the definition of a derivative. There is one small thing worth pointing out here because the natural logarithm of π₯ is only defined for positive values of π₯. We know that our definition for π prime of π₯ will also only be valid when π₯ is positive. Although this is not strictly necessary for how weβre going to use it, itβs worth keeping in mind.

The next piece of information we need is that π prime evaluated at one is equal to one. We can just find this by substituting π₯ is equal to one into our expression for π prime of π₯. And this is actually how weβre going to get our limit result. Remember, π prime of π₯ is the derivative of π of π₯ with respect to π₯. And we know how to define a derivative at a point by using limits. We need to recall the following definition of a derivative by using limits. For a constant π and differentiable function π of π₯, we define π prime of π to be equal to the limit as β approaches zero of π evaluated at π plus β minus π evaluated at π all divided by β provided that this limit exists. And we know this is what we mean when we say a function π is differentiable at a value of π. Weβre going to use this on the natural logarithm function at one. And in fact, we already know that this limit converges, and we know itβs equal to one. So, weβll start with one is equal to the derivative of the natural logarithm of π₯ with respect to π₯ evaluated at π₯ is equal to one.

Next, weβll use our definition for a derivative in terms of limits to write π prime of one as a limit. Itβs equal to the limit as β approaches zero of π evaluated at one plus β minus π evaluated at one all divided by β. And we know that this limit exists and it converges; itβs equal to one. Now what weβre going to do is rewrite this limit into a form which is very useful. The first thing weβre going to do is rewrite our variable β as the variable π₯. And although this step is not strictly necessary, most of the functions weβre going to be using this limit result on will be in terms of π₯. So, it makes sense to rewrite our limit result in terms of π₯. This gives us the limit as π₯ approaches zero of π of one plus π₯ minus π of one all divided by π₯.

But remember, we know that the function π of π₯ is just the natural algorithm function, so we can evaluate π at one plus π₯ and π at one. Doing this, we get the limit as π₯ approaches zero of the natural logarithm of one plus π₯ minus the natural logarithm of one all divided by π₯. And of course, we can simplify this. We can evaluate the natural logarithm of one. We know that π to the zeroth power is equal to one. This means the natural logarithm of one is just equal to zero. So, in actual fact, we can just remove this part from our limits altogether. And there is one more thing weβll do to simplify this limit. Instead of dividing by π₯, weβre going to multiply by one over π₯.

So, weβve now rewritten our limit as the limit as π₯ approaches zero of one over π₯ times the natural logarithm of one plus π₯. And we can simplify this limit even further. We need to notice something; weβre multiplying a logarithmic function by another function. And this should remind us of the power rule for logarithms. In terms of the natural logarithm, this tells us that π times the natural logarithm of π is equal to the natural logarithm of π raised to the power of π. So, inside of our limit, instead of multiplying by one over π₯ inside of our logarithm, we can raise one plus π₯ to the power of one over π₯. So, weβve now rewritten our limit as the limit as π₯ approaches zero of the natural logarithm of one plus π₯ all raised to the power of one over π₯. It can help to add an extra set of parentheses in here to remind us that weβre raising one plus π₯ to the power of one over π₯.

We need one more piece of simplification to get to our limit result. We can now see weβre taking the natural logarithm of some value inside of our limit. But we know something very interesting about the natural logarithm function. We know that the natural logarithm function is a continuous function. In fact, itβs continuous for all values of π₯ greater than zero. So, we have the limit of the natural logarithm of some function as π₯ approaches zero is equal to one. Remember, we know that this limit converges. So because the natural logarithm function is continuous and this limit evaluates to give us one, we can just take the natural logarithm function outside of our limit. So, by doing this, we now have the natural logarithm of the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯. And remember, weβve already shown this is equal to π prime of one which we know is equal to one.

And of course, we can simplify this expression. We can raise both sides as an exponent of π. If these two things are equal, then π to the first power should be equal to π to the power of the natural logarithm of the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯. And of course, we know that π and the natural logarithm function are inverse functions. So, this is just going to simplify to give us our limit. And then because π to the first power is just equal to π, weβve shown that π is equal to the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯. This is a very useful limit result.

For example, if we were to just try and evaluate our limit directly, we would see that when π₯ approaches zero, one plus π₯ is approaching one. However, one over π₯ is approaching β. So, trying to evaluate this limit directly, we get one to the power of β, which we know is an indeterminate form. So instead, when we get limits which look like this, we can try and rewrite them in terms of this limit so that our answer will be in terms of π.

Before we move on to see how we can use this limit result, we can actually rewrite this limit in another useful form. So, weβll clear some space and start with our limit result π is equal to the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯. To get our new useful limit result, weβre going to need to use a substitution. We want to set π₯ equal to one over π. If we do this, we can write the π₯ inside of our parentheses as one over π, giving us the limit as π₯ approaches zero of one plus one over π all raised to the power of one over π₯. However, we want to rewrite this entire limit in terms of π.

Next, weβre going to need to rewrite our exponent one over π₯. So, to do this, we need to find an expression for one over π₯. We can just take the reciprocal of both sides of our equation. If π₯ is equal to one over π, then we must have that one over π₯ is equal to π. So, we can replace the one over π₯ in our exponent with π, giving us the limit as π₯ approaches zero of one plus one over π all raised to the πth power. But our limit is still as π₯ is approaching zero, so we need to rewrite our limit in terms of π. And this is where we arrive at a small problem. We want to know what happens to our value of π as π₯ approaches zero. So, we want to use the equation one over π₯ is equal to π.

However, as π₯ approaches zero, we have several different options about what can happen to one over π₯. For example, the limit as π₯ approaches zero from the right of one over π₯ will be equal to positive β. Weβre dividing a positive number by a smaller and smaller positive number. This is going to grow without bound. However, the opposite is also true. We can have π₯ approaches zero from the left. Then, the limit as π₯ approaches zero from the left of one over π₯ will be equal to negative β. So, it seems like we donβt know what will happen to the value of π as π₯ approaches zero. However, itβs actually possible to sidestep this issue altogether.

If we go back to our original limit result, weβve already proven that this is true. π is equal to the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯. And if this limit is equal to π, then its left and right limit must be equal and must also be equal to π. In other words, π is also equal to the limit as π₯ approaches zero from the right of one plus π₯ all raised to the power of one over π₯.

So, when weβre doing these calculations, instead of taking π₯ approaches zero, we can actually just take π₯ approaches zero from the right. Then, we know as π₯ approaches zero from the right, our values of π are going to approach β. So, when π₯ approaches zero from the right, π is approaching β. So, weβve shown the limit as π approaches β of one plus one over π all raised to the πth power is also equal to π. And this gives us our second limit result π is equal to the limit as π approaches β of one plus one over π all raised to the πth power. These limit results are actually just restatements of each other; however, seeing them in different forms can be helpful for different situations. And itβs also worth pointing out sometimes youβll see the top limit result written in terms of π, and sometimes youβll see the bottom limit result written in terms of π₯. These are just variables, and it doesnβt matter what we call them.

And before we move on to seeing how we can use this limit result, weβll sketch a graph of π¦ is equal to one plus π₯ all raised to the power of one over π₯. If we plot this curve for values of π₯ greater than negative one, weβll get a sketch which looks somewhat like the following. We have a vertical asymptote when π₯ is equal to negative one and a horizontal asymptote at π¦ is equal to one. And we also know that our curve is not defined when π₯ is equal to zero because then we would have one divided by zero in our function. This is represented by the hollow circle in our diagram. If we were to plot this accurately, we would see that its position is at π. Letβs now move on to some examples to see how weβll use these two limit results.

Determine the limit as π₯ approaches β of one plus one over π₯ all raised to the power of four π₯.

Weβre given the limit which we need to evaluate. And the first thing we can try and do is just evaluate this limit directly. We see that our limit is as π₯ is approaching β. And we know as π₯ approaches β, the reciprocal function one over π₯ approaches zero. So, the inner part of our parentheses are approaching zero. However, as π₯ approaches β, four π₯ is approaching β. So, trying to evaluate this limit directly, we get one to the power of β, and this is an indeterminate form. So, weβre going to need to try some other method of evaluating this limit. Instead, we need to notice the limit given to us in the question is very similar to the limit in the definition of Eulerβs number. So, we need to recall the following limit result.

We know the limit as π₯ approaches β of one plus one over π₯ all raised to the power of π₯ is equal to Eulerβs constant π, and itβs worth committing this limit to memory. And itβs also worth pointing out sometimes youβll see this written in terms of the variable π. However, weβre given the limit in terms of π₯, so weβve just rewritten this in terms of π₯. And we can see the limit given to us in the question is almost exactly in this form. However, in our exponent, instead of just having π₯, we have four π₯. So, we want to rewrite this limit into a form where we can just use our limit result. And to do this, weβre going to need to use our laws of exponents.

First, we need to recall π to the power of π times π is equal to π to the power of π all raised to the power of π. And we want to use this to rewrite our limit into a form where we can use our limit result. This means weβre first going to want to rearrange the two factors in our exponent. By doing this and then using π is equal to one plus one over π₯, π is equal to π₯, and π is equal to four, weβve rewritten our limit as the limit as π₯ approaches β of one plus one over π₯ all raised to the power of π₯ all raised to the power of four.

But we canβt yet use our limit result because weβre raising this all to the fourth power. So, weβre going to want to apply the power rule for limits. One version of this tells us for any integer π and real constant π such that the limit as π₯ approaches π of π of π₯ exists, then the limit as π₯ approaches π of π of π₯ raised to the πth power is equal to the limit as π₯ approaches π of π of π₯ all raised to the πth power. And itβs worth pointing out this is also true if weβre taking limits at β. We want to apply this with π set to be four, π equal to β, and π of π₯ to be one plus one over π₯ all raised to the power of π₯.

And Itβs worth pointing out we know the limit as π₯ approaches β of π of π₯ exists because itβs equal to π. So, by the power rule for limits, we can rewrite our limit as the limit as π₯ approaching β of one plus one over π₯ all raised to the power of π₯, and then we raise this limit result to the power of four.

And now, we can just evaluate our inner limit. We know what itβs equal to; itβs equal to Eulerβs constant π. So, by replacing this limit with π, we get π to the fourth power, which is our final answer. Therefore, we were able to show the limit as π₯ approaches β of one plus one over π₯ all raised to the power of four π₯ is just equal to π to the fourth power.

Letβs now see how we can use our other limit result to help us evaluate a different limit.

Determine the limit as π₯ approaches zero of π₯ plus one all raised to the power of 11 over 10π₯.

Weβre asked to evaluate a limit, and we can try and do this directly. First, we can see our values of π₯ are approaching zero. This means inside of our parentheses, one plus π₯ is approaching one plus zero, which is one. However, we then see a problem when we try to evaluate our exponent. The numerator remains constant. However, as π₯ is approaching zero, the denominator is approaching zero, so the size of our exponent is growing without bound. This means by trying to evaluate our limit, we get an indeterminate form. Weβre going to need to try some other method of evaluating this limit.

To do this, we can take a closer look to the limit given to us in the question. Itβs actually very similar to a limit which we do know how to evaluate. We can see the limit given to us in the question is very similar to a useful limit result, the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯ is equal to Eulerβs constant π. Weβre going to want to rewrite the limit given to us in the question in a form where we can use our limit result.

The first thing weβre going to want to do is reorder the two terms inside of our parentheses. This is just so it matches the limit result weβve used. This gives us the limit as π₯ approaches zero of one plus π₯ all raised to the power of 11 over 10π₯. And now, we can see that this limit is very similar to our limit result. However, in our limit result, the exponent is one over π₯. However, in our limit, the exponent is 11 divided by 10π₯. So, we want to rewrite our exponent in terms of one over π₯. To do this, weβre going to start by writing 11 over 10π₯ as one over π₯ multiplied by 11 over 10. This gives us the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯ times 11 over 10.

Now, we want to write this in terms of our limit result. To do this, we need to do two things. First, weβre going to need to use our laws of exponents. First, we need to recall that π to the power of π times π can be rewritten as π to the power of π all raised to the power of π. We can use this to rewrite our limit with π set to be one plus π₯, π set to be one over π₯, and π set to be 11 over 10. Doing this, we get the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯ all raised to the power of 11 over 10. But we still canβt use our limit result because weβre raising this to the power of 11 over 10. We want to take this outside of our limit.

And in fact, we can do this in this case by using the power rule for limits. We recall this tells us for constant π, rational number π, and function π of π₯, the limit as π₯ approaches π of π of π₯ raised to the πth power is equal to the limit as π₯ approaches π of π of π₯ all raised the πth power. And this version of the power rule for limits will work, provided the limit as π₯ approaches π of π of π₯ exists and is equal to a nonnegative number.

And in fact, this is exactly the case we have. We know the limit as π₯ approaches zero of one plus π₯ all raised to the power of one over π₯ is equal to Eulerβs constant π. So, by using the power rule for limits, we can take 11 over 10 outside of our limit. And this just means we can evaluate the limit inside of our parentheses. We know this is equal to Eulerβs constant π. So, by writing this limit as π, we were able to show the limit as π₯ approaches zero of π₯ plus one all raised to the power of 11 over 10π₯ is equal to π to the power of 11 over 10.

Letβs now take a look at an example which will take more manipulation to get it in a form which we can evaluate.

Determine the limit as π₯ approaches β of one minus seven over π₯ all raised to the power of five π₯.

In this question, weβre asked to evaluate a limit. And we might be tempted to try and doing this directly. And if we were to try and do this directly, we would get a problem. Negative seven over π₯ as π₯ is approaching β is approaching zero and five π₯ is approaching β. So, this limit is approaching one to the power of β, which we know is an indeterminate form. So, weβre going to need to try a different method to evaluate this limit. In fact, there are two different methods to evaluate this limit. Weβll only go through one of them.

We know that the limit result given to us the question is very similar to our limits involving Eulerβs number π. And in fact, we could use either of these two definitions to evaluate our limit. Itβs personal preference which one youβd use. However, usually one of the two limits is easier than the other one. And itβs very difficult to see just from the limit youβre asked which one to use. So, if you get stuck using one of the definitions, try using the other one. So, we want to try and rewrite our limit in the following form. And, of course, we can immediately see some problems.

First, inside of our parentheses, instead of having π, we have negative seven over π₯. So, to get around this problem, weβre just going to try setting π equal to negative seven over π₯. We then want to use this substitution to rewrite our limit. Weβll start by rewriting negative seven over π₯ as π. This gives us the limit as π₯ approaches β of one plus π all raised to the power of five π₯. Of course, we want to rewrite this limit in terms of π. So, letβs find an expression for five π₯. And to do this, we can just rearrange our expression π is equal to negative seven over π₯. We can multiply both sides of this equation through by π₯ and then divide through by π. We get π₯ is equal to negative seven over π. And we can then substitute this into our limit.

This gives us the limit as π₯ approaches β of one plus π all raised to the power of five times negative seven over π. And in fact, we can simplify our exponent. Itβs equal to negative 35 over π. So, we now need to evaluate the limit as π₯ approaches β of one plus π all raised to the power of negative 35 over π. However, this is a problem. We have π₯ approaching β. We want to know what happens to π. To find this out, we remember that we set π equal to negative seven over π₯. So, we can just ask the question βwhat happens to π as π₯ approaches β?β As π₯ is approaching β, the denominator of this expression is growing without bound. So, π is a negative number with smaller and smaller magnitude; π is approaching zero from the left. So, we can just rewrite this as the limit as π approaches zero from the left of one plus π all raised to the power of negative 35 over π.

And itβs worth pointing out we donβt technically need the fact that π approaches zero from the left. We can just write this as π approaches zero. It doesnβt actually change the value of this limit. And now, this is almost in a form we can evaluate by using our limit result. All, we need to do is write this in terms of being to the power of one over π. First, weβll rewrite our limit by using our laws of exponents. Itβs equal to the limit as π approaches zero of one plus π all raised to the power of one over π all raised to the power of negative 35.

Now, all we need to do is take this exponent of negative 35 outside of our limit. And to do this, we will use the power rule for limits. The limit as π approaches π of π of π raised to the πth power is equal to the limit as π approaches π of π of π all raised to the power of π. And this is true provided π as an integer and the limit as π approaches π of π of π exists. And in this case, we know itβs true. We know itβs equal to Eulerβs constant π. Therefore, we can just take the exponent of negative 35 outside of our limit. Then, we can just evaluate this limit to give us π. So, our answer was π to the power of negative 35. And we can rewrite this as our final answer: one over π to the 35th power.

We can now go over the key points of this video. We found two useful limit results, and we showed that we can use these results to evaluate other limits by using algebraic manipulation, substitution, and the power rule for limits.