Euler's Number

Euler's Number e is a fundamental constant approximately equal to 2.71828 that models continuous exponential growth and decay.

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

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What each symbol means

What Euler's Number takes
ee
nn
Euler's Number
SymbolMeaning
eeEuler's Number, a mathematical constant approximately equal to 2.71828; it is the unique base of the natural logarithm and the value that the limit expression converges to as nn approaches infinity.
nnA positive integer representing the number of compounding periods or the parameter in the limit; as nn increases toward infinity, the expression (1+1n)n\left(1 + \frac{1}{n}\right)^{n} approaches ee more closely.

When to use it

Use Euler's Number when you encounter continuous compound interest, exponential growth and decay, or see the limit definition limn(1+1n)n\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n}.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Evaluate the limit expression for increasing values of n

Problem

Calculate (1+1n)n\left(1 + \frac{1}{n}\right)^{n} for n=50n = 50 to see how the result approaches e2.71828e \approx 2.71828.
  1. (1+150)50\left(1 + \frac{1}{50}\right)^{50}

    Substitute n=50n = 50 into the limit expression.

  2. =(1.02)50= \left(1.02\right)^{50}

    Evaluate the fraction: 1+150=1.021 + \frac{1}{50} = 1.02.

  3. 2.6916\approx 2.6916

    Use a calculator to compute (1.02)50(1.02)^{50}; the result is close to but slightly less than e2.71828e \approx 2.71828.

Answer: (1+150)502.6916\left(1 + \frac{1}{50}\right)^{50} \approx 2.6916

This example directly demonstrates the limit definition. Notice that n=50n = 50 gives a value already very close to ee; if we used n=1000n = 1000 or n=10000n = 10000, we would get even closer. You are witnessing the limit in action—as nn grows, the expression converges to exactly ee.

2. Exponential decay using negative exponents

Problem

A medication leaves the body exponentially according to N(t)=N0ektN(t) = N_0 e^{-kt}, where N0=500N_0 = 500 mg is the initial dose, k=0.2k = 0.2 is the decay constant per hour, and tt is time in hours. How much medication remains after 4 hours?
  1. N(4)=500e0.24N(4) = 500 \cdot e^{-0.2 \cdot 4}

    Substitute the values: N0=500N_0 = 500, k=0.2k = 0.2, and t=4t = 4 into the exponential decay formula.

  2. =500e0.8= 500 \cdot e^{-0.8}

    Multiply the exponent: 0.2×4=0.8-0.2 \times 4 = -0.8.

  3. =5000.4493224.65= 500 \cdot 0.4493 \approx 224.65

    Calculate e0.80.4493e^{-0.8} \approx 0.4493 using a calculator, then multiply by 500.

Answer: N(4)224.65 mgN(4) \approx 224.65 \text{ mg}

This problem shows why ee is essential in modeling decay. The negative exponent represents the loss of medication over time, and ee is the only base where the mathematical model matches how real biological processes actually work. Using ee instead of another base like 2 or 10 makes the rate constant kk have a direct, meaningful interpretation in the real world.

3. Newton's Law of Cooling in a laboratory experiment

Problem

In a physics lab, a cup of hot water cools according to T(t)=Troom+(T0Troom)ektT(t) = T_{\text{room}} + (T_0 - T_{\text{room}})e^{-kt}, where Troom=22°CT_{\text{room}} = 22°C is the room temperature, T0=100°CT_0 = 100°C is the initial water temperature, k=0.15k = 0.15 per minute is the cooling rate, and tt is time in minutes. What is the water's temperature after 6 minutes?
  1. T(6)=22+(10022)e0.156T(6) = 22 + (100 - 22) e^{-0.15 \cdot 6}

    Substitute all values into Newton's Law of Cooling: room temperature 22°C, initial temperature 100°C, cooling constant 0.15 per minute, and time 6 minutes.

  2. =22+78e0.9= 22 + 78 \cdot e^{-0.9}

    Simplify: 10022=78100 - 22 = 78 for the temperature difference, and 0.15×6=0.9-0.15 \times 6 = -0.9 for the exponent.

  3. =22+780.406622+31.71=53.71= 22 + 78 \cdot 0.4066 \approx 22 + 31.71 = 53.71

    Evaluate e0.90.4066e^{-0.9} \approx 0.4066 using a calculator, multiply by 78 to get 31.71, then add the room temperature.

Answer: T(6)53.71°CT(6) \approx 53.71°C

This realistic scenario shows how Euler's Number emerges naturally in physics. The term ekte^{-kt} captures the water's exponential approach toward room temperature: the water cools quickly at first (when the temperature difference is large), then more slowly as it approaches the room's temperature. Scientists and engineers use ee as the base precisely because this exponential model matches real-world heat transfer.

Common mistakes

Where Euler's Number usually goes wrong
Answer came out wrong
Writing e2x=2exe^{2x} = 2e^x or claiming (ex)2=2ex(e^x)^2 = 2e^x.
Use the exponent rule (am)n=amn(a^m)^n = a^{mn}: (ex)2=ex2=e2x(e^x)^2 = e^{x \cdot 2} = e^{2x}. Also remember ea+b=eaebe^{a+b} = e^a \cdot e^b, not ea+ebe^a + e^b.
Confusing the limit limn(1+1n)n\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} with something like (1+1n)n\left(1 + \frac{1}{n}\right) \cdot n or 1+1nn1 + \frac{1}{n^n}.
Remember the structure: (1+small)big(1 + \text{small})^{\text{big}} where small =1n= \frac{1}{n} and big =n= n. As nn increases, the base approaches 1 but is raised to a higher power, and these two competing effects balance perfectly to converge to ee.
Approximating e2.7e \approx 2.7 or e3e \approx 3 too early and carrying that approximation through multiple calculation steps.
Keep ee symbolic or use more decimal places (e2.71828e \approx 2.71828) until the final step, especially if the exponent is large. Only substitute a numerical value at the end when you report your answer.
The mistakeWhy it is wrongThe fix
Writing e2x=2exe^{2x} = 2e^x or claiming (ex)2=2ex(e^x)^2 = 2e^x.Exponential functions follow exponent rules, not linear rules; (ex)2=e2x(e^x)^2 = e^{2x}, not 2ex2e^x.Use the exponent rule (am)n=amn(a^m)^n = a^{mn}: (ex)2=ex2=e2x(e^x)^2 = e^{x \cdot 2} = e^{2x}. Also remember ea+b=eaebe^{a+b} = e^a \cdot e^b, not ea+ebe^a + e^b.
Confusing the limit limn(1+1n)n\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} with something like (1+1n)n\left(1 + \frac{1}{n}\right) \cdot n or 1+1nn1 + \frac{1}{n^n}.The exponent nn applies to the entire quantity (1+1n)\left(1 + \frac{1}{n}\right); the base and exponent grow together in a specific way, and only this pairing produces ee.Remember the structure: (1+small)big(1 + \text{small})^{\text{big}} where small =1n= \frac{1}{n} and big =n= n. As nn increases, the base approaches 1 but is raised to a higher power, and these two competing effects balance perfectly to converge to ee.
Approximating e2.7e \approx 2.7 or e3e \approx 3 too early and carrying that approximation through multiple calculation steps.In exponential problems, small errors in ee compound rapidly because of the exponent; a 0.01 error in ee becomes much larger when raised to a power.Keep ee symbolic or use more decimal places (e2.71828e \approx 2.71828) until the final step, especially if the exponent is large. Only substitute a numerical value at the end when you report your answer.

Tips and when to use something else

  • Look for the word 'continuous' in word problems—continuous compound interest, continuous growth, continuous decay. When you see 'continuous,' use exponential functions with base ee (like A=PertA = Pe^{rt}), not discrete compounding formulas like A=P(1+r)nA = P\left(1 + r\right)^n.
  • The derivative of exe^x is exe^x itself, which is why ee is special in calculus and why it appears whenever a quantity's growth rate is proportional to the amount present (bacteria, radioactivity, money in an account).
  • If you don't have ee memorized, remember that e12.7e^1 \approx 2.7 and e27.4e^2 \approx 7.4, so ee is between 2 and 3. For precision, use e2.71828e \approx 2.71828, and consult a table or calculator for values like e0.51.649e^{0.5} \approx 1.649 or e10.368e^{-1} \approx 0.368.
  • Do not confuse 'natural' in 'natural logarithm' or 'natural exponential' with 'simple'—'natural' means ee is the base, the one that appears most often in calculus and nature, not the one that's easiest to compute by hand.

Frequently asked questions

Why does Euler's Number come from a limit with nn \to \infty?
The limit represents the outcome of compounding interest more and more frequently without bound. As you increase the number of compounding periods toward infinity, you approach the maximum possible return—that limit is ee. This is why ee describes continuous processes: it is the natural result of letting discrete growth become infinitely frequent.
Is Euler's Number the same as the 'e' in scientific notation?
No—in scientific notation, 'ee' means 'times 10 to the power' (for example, 3.5e2=3.5×102=3503.5e^2 = 3.5 \times 10^2 = 350). Euler's Number is a specific constant, approximately 2.71828, which is the base of the natural logarithm. Context makes it clear which one is meant, but they are completely different.
Can I use 2 or 10 instead of ee for exponential functions?
You can, and for some calculations 2 or 10 is easier to compute by hand. However, ee is special: it is the only base where the derivative of axa^x equals axa^x itself. This makes ee the 'natural' base for modeling real-world processes where growth rate is proportional to the current amount, which is why it dominates in radioactive decay, population biology, and finance.
What is ln(e)\ln(e), and why is it important?
ln(e)=1\ln(e) = 1 by definition, because the natural logarithm is the inverse of the exponential function with base ee. Just as log10(10)=1\log_{10}(10) = 1, we have ln(e)=1\ln(e) = 1. This is one of many reasons ee is so useful: the mathematics involving it is simpler and cleaner than with other bases.

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Reviewed 2026-09-18