Limits Involving the Definition of e, Power Expansions & Higher Derivatives of Inverse Functions

Published

October 6, 2025

The number \(e\) keeps popping up everywhere in math, and today you’ll see why. We’ll tackle limits that look impossible at first – expressions like \((1 + \text{tiny number})^{\text{huge power}}\) – and learn a powerful logarithm trick that cracks them wide open. You’ll also discover a magical shortcut: if you know a function’s power series, you can read off its 2024th derivative at zero without differentiating even once. Seriously.

Why Limits Involving e and Power Expansions Matter

The number \(e\) and the techniques for evaluating tricky limits show up constantly in science and everyday life:

Finance: compound interest formulas use \((1 + r/n)^n\), which approaches \(e^r\) – this is exactly how banks calculate continuously compounded interest
Medicine: drug concentration in the bloodstream decays exponentially, and doctors use limits involving \(e\) to determine dosing schedules
Engineering: electrical circuits with capacitors charge and discharge following \(e\)-based exponential curves, and engineers approximate these with power expansions
Computer science: algorithms that divide problems in half have logarithmic complexity – understanding \(\ln(1+x)\) expansions helps analyze their performance
Physics: radioactive decay, heat transfer, and population dynamics all rely on exponential functions built from the definition of \(e\)

Today we’ll master techniques for evaluating limits that disguise the definition of \(e\), see when power expansion works (and when it fails!), and discover how Maclaurin series let us find absurdly high-order derivatives without differentiating.

Topics Covered

The definition of \(e\): \(\displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e\)
Variations of the limit definition: \(\displaystyle\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n = e^{-x}\)
Evaluating limits like \(\displaystyle\lim_{n \to \infty}\left(1 - \frac{3}{n^2}\right)^n\) using the logarithm technique
Re-deriving the power expansion of \(\ln(1+x)\) from first principles
Using \(\ln(1+x)\) expansions to handle limits where direct binomial expansion fails
Why binomial expansion of \((1 + 1/n)^n\) does not converge via truncation in powers of \(n\)
The Maclaurin coefficient formula: \(f^{(k)}(0) = k! \cdot a_k\)
Odd and even functions and their implications for Maclaurin expansions
Finding higher derivatives of \(\arcsin x\) using power series

Lecture Video

Key Frames from the Lecture

What You Need to Know First

What is the number e?

The number \(e \approx 2.71828\ldots\) is one of the most important constants in mathematics. It is defined as:

\[e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n\]

As \(n\) grows, the base \(1 + 1/n\) gets closer to 1, but the exponent \(n\) grows without bound. These two opposing forces balance out to produce the finite number \(e\). This is a \(1^\infty\) indeterminate form – the result could be 1, infinity, or anything in between, depending on how the base and exponent interact.

What is the Maclaurin expansion of ln(1+x)?

We proved in a previous lesson that:

\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^k}{k}\]

This comes from integrating the geometric series \(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\) term by term. The constant of integration is found by plugging in \(x = 0\) (since \(\ln 1 = 0\)).

What is a binomial expansion?

The binomial expansion generalizes \((1+x)^n\) for any exponent \(n\), even negative or fractional:

\[(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\]

When \(n\) is a negative integer like \(-1\), this gives us the geometric series:

\[(1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots\]

What are odd and even functions?

An even function satisfies \(f(-x) = f(x)\) – its graph is symmetric about the \(y\)-axis. Examples: \(x^2\), \(\cos x\).

An odd function satisfies \(f(-x) = -f(x)\) – its graph has rotational symmetry about the origin. Examples: \(x^3\), \(\sin x\).

In terms of Maclaurin series:

Even functions have only even powers of \(x\)
Odd functions have only odd powers of \(x\)

This means that for an odd function, all even-order derivatives at zero are automatically zero!

What is the relationship between Maclaurin coefficients and derivatives?

If a function has a Maclaurin expansion \(f(x) = \sum_{k=0}^{\infty} a_k x^k\), then each coefficient encodes a derivative:

\[a_k = \frac{f^{(k)}(0)}{k!} \quad \Longleftrightarrow \quad f^{(k)}(0) = k! \cdot a_k\]

So if you know the power series, you can read off any derivative at zero without differentiating!

Key Concepts

Variations on the Definition of e

We start from the fundamental definition:

\[\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e\]

The key feature is that the term added to 1 inside the parentheses is the reciprocal of the exponent. We can create variations by forcing other expressions into this same shape.

Example 1: Find \(\displaystyle\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n\).

We rewrite the expression to match the definition. Since we have \(-x/n\) inside and \(n\) as the exponent, we can write:

\[\left(1 + \frac{(-x)}{n}\right)^n\]

To force this into the \(e\)-definition form, we need the exponent to be the reciprocal of \(-x/n\), which is \(-n/x\). So we write:

\[\left(1 + \frac{(-x)}{n}\right)^{(-n/x) \cdot (-x)} = \left[\left(1 + \frac{(-x)}{n}\right)^{-n/x}\right]^{-x}\]

As \(n \to \infty\), the inner bracket approaches \(e\) by definition, so the limit is:

Key Idea: Variations on the Definition of e

Any limit of the form \((1 + \text{something}/n)^n\) can be reshaped into the definition of \(e\). The trick is to rewrite the expression so the exponent is the reciprocal of the small term inside the parentheses.

\[\boxed{\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n = e^{-x}}\]

Explore – see how \((1 + 1/n)^n\) converges to \(e\):

Drag the slider for \(n\) and watch the green dot approach the dashed red line \(y = e \approx 2.718\).

The Logarithm Technique for Limits

Example 2: Find \(\displaystyle\lim_{n \to \infty}\left(1 - \frac{3}{n^2}\right)^n\).

This one cannot be directly forced into the \(e\)-definition because the term inside is \(-3/n^2\) while the exponent is only \(n\) (not \(n^2\)). Let’s call the limit \(L\) and take the natural log:

\[\ln L = \lim_{n \to \infty} n \cdot \ln\left(1 - \frac{3}{n^2}\right)\]

Now we apply the Maclaurin expansion \(\ln(1+u) = u - \frac{u^2}{2} + \cdots\) with \(u = -3/n^2\):

\[\ln\left(1 - \frac{3}{n^2}\right) = -\frac{3}{n^2} - \frac{9}{2n^4} - \cdots\]

Multiplying by \(n\):

\[n \cdot \ln\left(1 - \frac{3}{n^2}\right) = -\frac{3}{n} - \frac{9}{2n^3} - \cdots\]

As \(n \to \infty\), every term vanishes. So \(\ln L = 0\), which means:

\[\boxed{L = e^0 = 1}\]

We can verify this makes intuitive sense: the base \(1 - 3/n^2\) approaches 1 faster than the exponent \(n\) grows, so the limit collapses to 1.

Alternative approach: Force it into the \(e\)-definition directly. Rewrite as:

\[\left(1 + \frac{(-3)}{n^2}\right)^{(-n^2/3) \cdot (-3/n)} = \left[\left(1 + \frac{(-3)}{n^2}\right)^{-n^2/3}\right]^{-3/n}\]

The inner bracket approaches \(e\), and the outer exponent \(-3/n \to 0\). So we get \(e^0 = 1\). Same answer, confirmed.

A Tricky Product Limit

Example 3: Find \(\displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{2n}\right)^{2n^2}\left(1 - \frac{1}{3n}\right)^{3n^2}\).

This is a product of two terms, each in the \(1^\infty\) indeterminate form. A natural first thought is to force each into the \(e\)-definition:

\[\left(1 + \frac{1}{2n}\right)^{2n} \to e \quad \text{and} \quad \left(1 - \frac{1}{3n}\right)^{3n} \to e^{-1}\]

But the actual exponents are \(2n^2\) and \(3n^2\), not \(2n\) and \(3n\). We can rewrite the first factor as:

\[\left[\left(1 + \frac{1}{2n}\right)^{2n}\right]^{n} \to e^n \to \infty\]

and the second as:

\[\left[\left(1 - \frac{1}{3n}\right)^{3n}\right]^{n} \to e^{-n} \to 0\]

So we get \(\infty \cdot 0\), an indeterminate form. We need a more careful approach.

Why binomial expansion fails here: If you try to expand \((1 + 1/(2n))^{2n^2}\) using the binomial theorem, the coefficients \(\binom{2n^2}{k}\) contain powers of \(n\) in the numerator that cancel with the powers of \(1/(2n)\) in the denominator. The terms do not shrink based on powers of \(1/n\) – they shrink based on \(1/k!\). This means you cannot truncate to a few terms; you need infinitely many to capture the limit.

The logarithm technique works: Set \(L\) equal to the limit and take the log:

\[\ln L = \lim_{n \to \infty}\left[2n^2 \ln\left(1 + \frac{1}{2n}\right) + 3n^2 \ln\left(1 - \frac{1}{3n}\right)\right]\]

Now expand each logarithm using \(\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots\):

\[\ln\left(1 + \frac{1}{2n}\right) = \frac{1}{2n} - \frac{1}{8n^2} + \cdots\]

\[\ln\left(1 - \frac{1}{3n}\right) = -\frac{1}{3n} - \frac{1}{18n^2} - \cdots\]

Multiplying through:

\[2n^2 \cdot \ln\left(1 + \frac{1}{2n}\right) = 2n^2\left(\frac{1}{2n} - \frac{1}{8n^2} + \cdots\right) = n - \frac{1}{4} + \cdots\]

\[3n^2 \cdot \ln\left(1 - \frac{1}{3n}\right) = 3n^2\left(-\frac{1}{3n} - \frac{1}{18n^2} - \cdots\right) = -n - \frac{1}{6} + \cdots\]

Adding them together:

\[\ln L = \left(n - \frac{1}{4} + \cdots\right) + \left(-n - \frac{1}{6} + \cdots\right) = -\frac{1}{4} - \frac{1}{6} + \cdots = -\frac{5}{12}\]

The \(n\) terms cancel (as expected), and the higher-order terms vanish. Therefore:

\[\boxed{L = e^{-5/12}}\]

This is a beautiful result – a number between 0 and 1 that you could never guess without the technique.

Explore – watch the product converge as \(n\) grows:

Increase \(n\) with the slider and watch the green dot converge to \(e^{-5/12} \approx 0.659\).

Why the Logarithm Technique is Essential

The key insight from the product limit example is this: when you have \((1 + \text{small})^{\text{large}}\) and try to expand directly with the binomial theorem, the convergence depends on \(1/k!\) (the factorial in the denominator), not on powers of \(1/n\). This means:

You cannot truncate the expansion at a few terms
The “lowest order infinitesimal” strategy does not apply directly

But when you take the natural log, the exponent comes down as a multiplier:

\[\ln\left[(1 + u)^m\right] = m \cdot \ln(1+u)\]

Now the expansion of \(\ln(1+u)\) does depend on increasing powers of \(u\), so truncation works perfectly. This is why the logarithm technique is so powerful for these limit problems.

Re-deriving the Power Expansion of ln(1+x)

This derivation was reviewed in the lecture from scratch. Here is the chain of reasoning:

Step 1: We need the derivative of \(\ln u\). Since \(\ln\) is the inverse of \(e^x\), we write \(y = \ln u\), so \(u = e^y\). Differentiating:

\[\frac{du}{dy} = e^y = u \quad \Longrightarrow \quad \frac{dy}{du} = \frac{1}{u}\]

So \(\frac{d}{du}\ln u = \frac{1}{u}\). Note: this reciprocal relationship between the derivative of a function and its inverse only works for the first derivative.

Step 2: By the chain rule, \(\frac{d}{dx}\ln(1+x) = \frac{1}{1+x}\).

Step 3: Expand \(\frac{1}{1+x} = (1+x)^{-1}\) using the binomial/geometric series:

\[\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots\]

Step 4: Since \(\ln(1+x)\) is the antiderivative of \(\frac{1}{1+x}\), integrate term by term:

\[\ln(1+x) = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]

Step 5: Determine \(C\) by plugging in \(x = 0\): since \(\ln(1) = 0\), we get \(C = 0\).

Key Idea: Power Series for ln(1+x)

This series comes from integrating the geometric series \(\frac{1}{1+x} = 1 - x + x^2 - \cdots\) term by term. It is one of the most useful expansions in calculus – the key weapon for cracking \(1^\infty\) limits.

\[\boxed{\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots}\]

Higher Derivatives via Maclaurin Coefficients

Recall the general Maclaurin expansion:

\[f(x) = \sum_{k=0}^{\infty} a_k x^k \quad \text{where} \quad a_k = \frac{f^{(k)}(0)}{k!}\]

We derive this by differentiating the power series \(k\) times and evaluating at \(x = 0\). Each differentiation peels off one power of \(x\) and multiplies by the corresponding index, until after \(k\) differentiations only the constant \(k! \cdot a_k\) survives.

Application to arcsine: What is the 2025th derivative of \(\arcsin x\) at \(x = 0\)?

First, observe that \(\sin x\) is an odd function, and so is its inverse \(\arcsin x\). An odd function’s Maclaurin expansion contains only odd powers of \(x\):

\[\arcsin x = a_1 x + a_3 x^3 + a_5 x^5 + \cdots\]

The 2025th derivative corresponds to the coefficient of \(x^{2025}\). Since 2025 is odd, this coefficient could be nonzero – you would need to find the full expansion to determine the answer.

But what about the 2024th derivative of \(\arcsin x\) at \(x = 0\)? Since 2024 is even, there is no \(x^{2024}\) term in the expansion of an odd function. Therefore:

\[\boxed{(\arcsin x)^{(2024)}\bigg|_{x=0} = 0}\]

No computation needed – the symmetry of the function tells us the answer instantly.

Explore – see that \(\arcsin x\) is an odd function:

Notice that \(\arcsin x\) (blue) is the reflection of \(\sin x\) (red dashed) across the line \(y = x\). Both are odd functions, passing through the origin with rotational symmetry.

Finding the Expansion of arcsin x (Homework Preview)

To find the actual Maclaurin expansion of \(\arcsin x\), apply the same strategy used for \(\ln(1+x)\):

Take the derivative: \(\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-1/2}\)
Expand \((1 - x^2)^{-1/2}\) using the binomial series
Integrate term by term to recover \(\arcsin x\)
The constant is zero since \(\arcsin(0) = 0\)

This was assigned as homework – try it yourself!

Cheat Sheet

What you want	Formula
Definition of \(e\)	\(\displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e\)
General variation	\(\displaystyle\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n = e^x\)
Power series for \(\ln(1+x)\)	\(\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\)
Logarithm technique	Set \(L = \lim\), compute \(\ln L\), then \(L = e^{\ln L}\)
Maclaurin coefficient formula	\(a_k = \frac{f^{(k)}(0)}{k!}\), so \(f^{(k)}(0) = k! \cdot a_k\)
Odd function rule	Odd functions have only odd powers in their Maclaurin series; all even derivatives at 0 are zero
Derivative of \(\ln u\)	\(\frac{d}{du}\ln u = \frac{1}{u}\) (proved via inverse of \(e^x\))
Inverse function derivative	\(\frac{dy}{dx} = \frac{1}{dx/dy}\) (first derivative only!)

The Big Strategy for \(1^\infty\) Limits

\[\text{Set } L = \lim \;\xrightarrow{\;\text{take}\;\ln\;}\; \ln L = \lim\, m \cdot \ln(1+u) \;\xrightarrow{\;\text{expand}\;}\; \text{keep lowest order in } u \;\xrightarrow{\;\text{evaluate}\;}\; L = e^{\ln L}\]