Limit Definition of e

The limit definition of e shows how the mathematical constant e arises from the growth of compound interest and exponential functions.

limx0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e

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What each symbol means

What Limit Definition of e takes
xx
ee
Limit Definition of e
SymbolMeaning
xxx is a dimensionless variable approaching 0 in both the base 1+x1+x and the exponent 1/x1/x; misreading one of these roles breaks the entire formula.
eee is the mathematical constant approximately 2.71828, the base of natural logarithms; confusing it with a different constant like 10 or with a variable ee used elsewhere corrupts the result.

When to use it

Use this definition when evaluating limits of the form (1+x)1/x(1 + x)^{1/x} or when proving properties of the exponential function.

Level

Usually taught in: Calculus I

Worked examples

1. Evaluate a limit with a scaled base

Problem

Evaluate limx0(1+2x)1/x\lim_{x \to 0} (1 + 2x)^{1/x}.
  1. Letu=2xLet \quad u = 2x

    We set up a substitution so that as x0x \to 0, we have u0u \to 0, and the base becomes 1+u1 + u.

  2. x=u2and1x=2ux = \frac{u}{2} \quad \text{and} \quad \frac{1}{x} = \frac{2}{u}

    We solve for xx and 1/x1/x in terms of uu to rewrite the exponent.

  3. (1+2x)1/x=(1+u)2/u=[(1+u)1/u]2(1 + 2x)^{1/x} = (1 + u)^{2/u} = [(1 + u)^{1/u}]^2

    After substitution, we factor the exponent so that (1+u)1/u(1 + u)^{1/u} is isolated, matching the limit definition form.

  4. limx0(1+2x)1/x=limu0[(1+u)1/u]2=e2\lim_{x \to 0} (1 + 2x)^{1/x} = \lim_{u \to 0} [(1 + u)^{1/u}]^2 = e^2

    Taking the limit as u0u \to 0, the expression (1+u)1/u(1 + u)^{1/u} approaches ee, so the entire limit is e2e^2.

Answer: e2e^2

The key insight is recognizing that (1+2x)1/x(1 + 2x)^{1/x} can be rewritten using substitution to match the standard form (1+u)1/u(1 + u)^{1/u}. When the exponent is a constant multiple of 1/u1/u like 2/u2/u, we can factor it out and apply the limit definition to the core (1+u)1/u(1 + u)^{1/u} term, then raise the result to the appropriate power.

2. Evaluate a limit with a negative coefficient

Problem

Evaluate limx0(13x)1/x\lim_{x \to 0} (1 - 3x)^{1/x}.
  1. Letu=3xLet \quad u = -3x

    We substitute so that the base (13x)(1 - 3x) becomes (1+u)(1 + u), which matches the standard form.

  2. x=u3and1x=3ux = -\frac{u}{3} \quad \text{and} \quad \frac{1}{x} = -\frac{3}{u}

    Solving for xx and 1/x1/x in terms of uu gives us the exponent in the new variable.

  3. (13x)1/x=(1+u)3/u=[(1+u)1/u]3(1 - 3x)^{1/x} = (1 + u)^{-3/u} = [(1 + u)^{1/u}]^{-3}

    After substitution, the exponent is 3/u-3/u, and factoring reveals that the negative sign is preserved.

  4. limx0(13x)1/x=limu0[(1+u)1/u]3=e3\lim_{x \to 0} (1 - 3x)^{1/x} = \lim_{u \to 0} [(1 + u)^{1/u}]^{-3} = e^{-3}

    The core limit is ee, and we raise it to the power of 3-3 to get e3e^{-3}, which is the reciprocal of e3e^3.

Answer: e3e^{-3}

When the base has a negative coefficient, the substitution must convert the minus into a plus, which introduces a negative factor in the exponent. The negative sign carries through to the final answer as a negative exponent, illustrating how the limit definition handles both growth (ek)(e^k) and decay (ek)(e^{-k}) patterns.

3. Model cooling behavior in a thermal lab

Problem

A thermal testing lab models the cooling of a heated object using the formula (10.2t)1/(0.2t)(1 - 0.2t)^{1/(0.2t)} where tt is time in seconds. As the measurement interval shrinks to 0, find the limiting cooling factor.
  1. Letu=0.2tLet \quad u = -0.2t

    We substitute to convert (10.2t)(1 - 0.2t) into the standard form (1+u)(1 + u).

  2. t=5uand10.2t=1ut = -5u \quad \text{and} \quad \frac{1}{0.2t} = -\frac{1}{u}

    Solving for tt and the reciprocal of 0.2t0.2t in terms of uu allows us to rewrite the exponent.

  3. (10.2t)1/(0.2t)=(1+u)1/u=[(1+u)1/u]1(1 - 0.2t)^{1/(0.2t)} = (1 + u)^{-1/u} = [(1 + u)^{1/u}]^{-1}

    After substitution and factoring, we isolate the (1+u)1/u(1 + u)^{1/u} term with a negative exponent.

  4. limt0(10.2t)1/(0.2t)=limu0[(1+u)1/u]1=e1=1e\lim_{t \to 0} (1 - 0.2t)^{1/(0.2t)} = \lim_{u \to 0} [(1 + u)^{1/u}]^{-1} = e^{-1} = \frac{1}{e}

    As u0u \to 0, the limit is e1e^{-1}, which is the reciprocal of ee, showing that the cooling factor approaches 1/e1/e.

Answer: e1 or 1ee^{-1} \text{ or } \frac{1}{e}

This problem shows how the limit definition of ee naturally appears in real-world cooling models. The fact that the limiting cooling factor is exactly 1/e1/e demonstrates that the mathematical constant ee is not arbitrary but emerges from fundamental exponential processes in nature. By recognizing the form and applying substitution strategically, abstract calculus connects directly to measurable physical phenomena.

Common mistakes

Where Limit Definition of e usually goes wrong
Answer came out wrong
limx0(1+x)1/x=1\lim_{x \to 0} (1 + x)^{1/x} = 1 (treating it like an indeterminate form that equals 1)
Recognize that 11^{\infty} is indeterminate and does not default to 1; the limit definition of ee proves that this limit equals exactly ee.
When evaluating limx0(1+2x)1/x\lim_{x \to 0} (1 + 2x)^{1/x}, directly writing (1+2x)1/x=e2(1 + 2x)^{1/x} = e^{2} without substitution.
Use substitution u=2xu = 2x to rewrite (1+2x)1/x=[(1+u)1/u]2(1 + 2x)^{1/x} = [(1 + u)^{1/u}]^2, then apply the definition to get e2e^2.
Confusing limx0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e with limn(1+1/n)n=e\lim_{n \to \infty} (1 + 1/n)^n = e and mixing up which form applies to your problem.
Match your substitution to the form in your problem: the first definition applies when x0x \to 0 with exponent 1/x1/x; the second applies when nn \to \infty with exponent nn.
The mistakeWhy it is wrongThe fix
limx0(1+x)1/x=1\lim_{x \to 0} (1 + x)^{1/x} = 1 (treating it like an indeterminate form that equals 1)This is a 11^{\infty} indeterminate form, but it does not collapse to 1; instead it has the specific non-trivial limit value of e2.718e \approx 2.718.Recognize that 11^{\infty} is indeterminate and does not default to 1; the limit definition of ee proves that this limit equals exactly ee.
When evaluating limx0(1+2x)1/x\lim_{x \to 0} (1 + 2x)^{1/x}, directly writing (1+2x)1/x=e2(1 + 2x)^{1/x} = e^{2} without substitution.The base is (1+2x)(1 + 2x), not (1+x)(1 + x), so you cannot immediately apply the definition; you must rewrite it in the standard form first.Use substitution u=2xu = 2x to rewrite (1+2x)1/x=[(1+u)1/u]2(1 + 2x)^{1/x} = [(1 + u)^{1/u}]^2, then apply the definition to get e2e^2.
Confusing limx0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e with limn(1+1/n)n=e\lim_{n \to \infty} (1 + 1/n)^n = e and mixing up which form applies to your problem.These are two different (but equivalent) definitions of ee; using the wrong substitution strategy or failing to recognize which form your problem takes leads to incorrect limits.Match your substitution to the form in your problem: the first definition applies when x0x \to 0 with exponent 1/x1/x; the second applies when nn \to \infty with exponent nn.

Tips and when to use something else

  • Look for the pattern (1+something)1/(same something)(1 + \text{something})^{1/(\text{same something})} and use substitution to match the standard form before applying the limit definition.
  • If your limit has the form (1+x)a(1 + x)^a where aa is a constant, do not use the limit definition of ee; instead just evaluate (1+x)a(1 + x)^a at x=0x = 0 to get 1a=11^a = 1.
  • Remember that the two definitions of ee are equivalent: limx0(1+x)1/x=limn(1+1/n)n=e\lim_{x \to 0} (1 + x)^{1/x} = \lim_{n \to \infty} (1 + 1/n)^n = e; choose whichever form makes your problem simpler.
  • When the exponent is a multiple of 1/x1/x like (1+x)5/x(1 + x)^{5/x}, factor it as [(1+x)1/x]5[(1 + x)^{1/x}]^5 before taking the limit to isolate the definition.

Frequently asked questions

What does the ee in e2.718e \approx 2.718 have to do with compound interest?
The limit definition shows that ee naturally arises when you compound interest infinitely often. If you invest 1 dollar at 100% annual interest compounded nn times per year, as nn \to \infty your return approaches ee dollars. This is why ee is the base of natural exponential growth and decay in finance, biology, and physics.
Why do I need x0x \to 0 in the exponent and not nn \to \infty?
Both are valid—the definitions limx0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e and limn(1+1/n)n=e\lim_{n \to \infty} (1 + 1/n)^n = e are equivalent. Use whichever substitution makes your problem simpler. If your problem naturally has a term going to 0, use the first; if it naturally has a term going to infinity, use the second.
Can I use L'Hopital's Rule instead of the limit definition of ee?
Yes. Since (1+x)1/x(1 + x)^{1/x} has the indeterminate form 11^{\infty} as x0x \to 0, you can rewrite it as eln(1+x)/xe^{\ln(1 + x)/x} and apply L'Hopital's Rule to the exponent. However, the limit definition of ee is more direct and shows why ee is defined this way in the first place.
What if the coefficient in the base is not 1, like (1+3x)1/x(1 + 3x)^{1/x}?
Use substitution to rewrite it in the standard form. If u=3xu = 3x, then (1+3x)1/x=(1+u)3/u=[(1+u)1/u]3e3(1 + 3x)^{1/x} = (1 + u)^{3/u} = [(1 + u)^{1/u}]^3 \to e^3. The key is ensuring both the base coefficient and the exponent coefficient match so you can apply the definition.

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Reviewed 2026-09-18