In calculus, there are many ways to evaluate(i.e., finding the actual value) a limit. There is not a preferred method over another, you have to learn all of them and choose the right one according to the limit you are trying to solve. In this guide, we will see how to solve limits using Euler’s number($e$). Before going further, I will assume that you already know the formal definition of a limit, how to solve them(at least the simpler one) and the basics of algebra.

Euler’s Number⌗

Euler’s number is an irrational(it cannot be represented as a ratio between two numbers), transcendental(it is not a solution of any polynomial with rational coefficient) number discovered by the swiss mathematician Jacob Bernoulli while studying financial problems. The symbol $e$ is retained to Leonhard Euler. Along with $\pi$, it is one of the most common mathematical constant; it looks like this:

$$e = 2.7182818284590452353602874 \dots$$

Euler’s number $e$ can be defined with the following sequence: $$a_n := \left(1 + \frac{1}{n} \right)^n$$

or by using the following series: $$e := \sum_{n=0}^{+\infty} \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \dots + \frac{1}{n!}$$

Proving The Existence Of $e$⌗

We will give a proof only to the first definition. To do this, we will prove that the limit exists and that it is finite. Let us start by proving that the limit exists using Newton’s binomial:

$$a_n = \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} =$$

$$= \sum_{k=0}^{n} \frac{1}{k!} \frac{n(n-1)\dots (n-k+1)}{n^k}$$

$$= \sum_{k=0}^n \frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n-1}{n} \dots \frac{n-k+1}{n}$$

$$= \sum_{k=0}^n \frac{1}{k!} \cdot 1 \cdot \left( 1 - \frac{1}{n} \right) \dots \left( 1 - \frac{k-1}{n} \right)$$

Now let’s try to calculate the $a_{n=1}$ term of the sequence using the same formula: $$a_{n+1} = \sum_{k=0}^{n+1} \frac{1}{k!} \cdot 1 \cdot \left( 1 - \frac{1}{n+1} \right) \dots$$

$$\left( 1 - \frac{k-1}{n+1} \right)$$

$$\because \left( 1 - \frac{1}{n} \right) < \left( 1 - \frac{1}{n+1} \right) , \dots ,$$

$$\left( 1 - \frac{k-1}{n} \right) < \left( 1 - \frac{k-1}{n+1} \right)$$

then, $a_n < a_{n+1}$. The sequence $a_n$ is monotonic so the limit indeed exists. For the last part, let’s prove that the limit is finite by proving that the sequence is limited. That is: $$3 < a_n < 2 \qquad \forall n > 1$$

Since $a_2 = 1$ and $a_n > 2$ $\forall n > 1$(the sequence is monotonic) we now prove that $a_n < 3$ $\forall n > 1$. To do that we can use once again Newton’s formula. Furthermore we can observe that $$k! = 1 \cdot 2 \cdot 3 \dots k \geq 1 \cdot 2 \cdot 2 \dots 2 = 2^{k-1}$$

Thus, we have $$a_n = \sum_{k=0}^{n} \frac{1}{k!} \cdot 1 \cdot \left( 1 - \frac{1}{n} \right) \dots \left( 1 - \frac{k-1}{n} \right)$$

$$< \sum_{k=0}^n \frac{1}{k!}$$

$$= 1 + \sum_{k=1}^{n} \frac{1}{k!} \geq 1 + \sum_{k=1}^n \frac{1}{^{k-1}} = 1 + \sum_{k=0}^{n-1} \frac{1}{^k}$$

The last sum identifies the $n-$term of the sequence of partial sum of the geometric series with $q = \frac{1}{2}$. This led us to

$$\sum_{k=0}^{n-1} q^k = \frac{1-q^n}{1-q}$$

$$\iff \sum_{k=0}^{n-1} \frac{1}{2^k} = \frac{1-\frac{1}{2^k}}{1-\frac{1}{2}} = 2\left( 1 - \frac{1}{2^n} \right)$$

Which confirms that $a_n < 3$.

Applications of $e$⌗

There are many usage of Euler’s number on mathematics. For instance, it can be used to define the exponential function $f(x) = e^x$ or to define the natural logarithm $\ln(x) \equiv \log_e(x)$. Another cool application of $e$ involves complex numbers: $$e^{i \theta} = \cos(\theta) + i \sin(\theta) \ \ \forall \theta \in \mathbb{R}$$

The so-called Euler’s formula.

The other application of Euler’s number is the topic of this article.

Solving Limits With $e$⌗

From the definition of Euler’s number we know that $$\lim_{x \to \pm \infty} \left( 1 + \frac{1}{x} \right)^x = e$$

This means that $$\lim_{x \to \pm \infty} \left( 1 + \frac{1}{f(x)} \right)^{f(x)} = e \iff f(x) \to \pm \infty$$

Thus, if we have an indeterminate form that can be rewritten in this way, we can use our prior knowledge about this limit to easily calculate the first one. The best way to understand this, is the direct approach(i.e., by solving problems).

Example 1⌗

$$\lim_{x \to \infty} \left( 1 + \frac{3}{x} \right)^x$$

We can easily see(by substitution) that the limit is an indeterminate form $[1^{\infty}]$, but since $f(x) = \frac{x}{3} \to \infty$ we can use Euler’s number to get its value.

First, let’s rewrite our limit like this

$$\lim_{x \to \infty} \left( 1 + \frac{1}{\displaystyle \frac{x}{3}} \right)^x$$

Now we need to move the “3” to the exponent, to do that we can multiply and divide the exponent by “3”:

$$\lim_{x \to \infty} \left[ \left( 1 + \frac{1}{\displaystyle \frac{x}{3}} \right)^{x \cdot \displaystyle \frac{3}{3}} \right]$$

$$= \lim_{x \to \infty} \left[ \underbrace{\left( 1 + \frac{1}{\displaystyle \frac{x}{3}} \right)^{\displaystyle \frac{x}{3}}}_{\displaystyle e} \right]^3 = e^3$$

Example 2⌗

$$\lim_{x \to \infty} \left( \frac{3x+1}{3x-5} \right)^{x-2}$$

Again, this is an indeterminate form $[1^{\infty}]$. The first thing we can do to obtain the standard form is to add and subtract a quantity to the numerator of the fraction. In this case this quantity is $5$: $$\lim_{x \to \infty} \left( \frac{3x - 5 + 5 + 1}{3x-5} \right)^{x-2}$$ We can now split the fraction: $$\lim_{x \to \infty} \left( \frac{\cancel{3x-5}^1}{\cancel{3x-5}^1} + \frac{6}{3x-5} \right)^{x-2}$$

$$= \lim_{x \to \infty} \left( 1 + \frac{6}{3x-5} \right)^{x-2}$$

At this point we need to “adjust” our fraction in order to have $f(x) \to \infty$. To do so, we can rewrite it like this $$\lim_{x \to \infty} \left( 1 + \frac{1}{\displaystyle \frac{3x+5}{6}} \right)^{x-2}$$

Finally, let’s multiply and divide the exponent by $f(x)$:

$$\lim_{x \to \infty} \left( 1 + \frac{1}{ \frac{3x+5}{6}} \right)^{(x-2) \cdot \frac{3x+5}{6} \cdot \frac{6}{3x+5}}$$

which led us to $$\lim_{x \to \infty} \left[ \left( 1 + \frac{1}{\displaystyle \frac{3x+5}{6}} \right)^{\displaystyle \frac{3x+5}{6}} \right]^{\displaystyle \frac{6x-12}{3x+5}}$$

$$= e^{\displaystyle \frac{6}{3}} = e^2$$

Example 3⌗

$$\lim_{n \to \infty} \left( \frac{n^2 - 5n + 6}{n^2 - 9n + 20} \right)^{2n-1}$$

In this case, we first need to divide the numerator by the denominator(i.e., polynomial division). This leave us with two new polynomials: $$P(x) = 1 \qquad R(x) = 4n - 14$$

We can rewrite the limit like this: $$\lim_{n \to \infty} \left( 1 + \frac{4n - 14}{n^2 - 9n + 20} \right)^{2n-1}$$

From now on, it’s exactly like before.

$$= \lim_{n \to \infty} \left( 1 + \frac{1}{ \frac{n^2 - 9n + 20}{4n-14}} \right)^{ 2n-1 \cdot \frac{n^2 - 9n + 20}{4n - 14} \cdot \frac{4n - 14}{n^2 - 9n + 20}}$$

$$= \lim_{n \to \infty} \left[ \underbrace{\left( 1 + \frac{1}{ \frac{n^2 - 9n + 20}{4n - 14}} \right)^{ \frac{n^2 - 9n + 20}{4n - 14}}}_{ e} \right]^{ \frac{(4n-14)(2n-1)}{n^2 - 9n + 20}}$$

$$= e^8$$