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Find lim x->0 (1 + x)^(1/x): animated lesson

A narrated, step-by-step limits video lesson with captions. Final answer: Limit = e ≈ 2.718281828.

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (1.1 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
$\lim_{x \to 0} {\left(1+ x\right)}^{\frac{1}{x}}$
2. Steps, part 1
1. $\lim_{x \to 0} {\left(1+ x\right)}^{\frac{1}{x}} = e^{\,\lim_{x \to 0} \left[\frac{1}{x} \ln\left(1+ x\right)\right]}$
  Indeterminate form 1^∞ — take logarithms
2. $\lim_{x \to 0} \frac{1}{x} \ln\left(1+ x\right) = \lim_{x \to 0} \frac{\ln\left( x+1\right)}{x}$
  Indeterminate form 0·∞ — rewrite as a quotient
3. $\lim_{x \to 0} \frac{\ln\left( x+1\right)}{x} = \lim_{x \to 0} \frac{\frac{1}{x+1}}{1}$
  Indeterminate form 0/0 — apply L'Hôpital's rule
4. $\lim_{x \to 0} \frac{\frac{1}{x+1}}{1} = \lim_{x \to 0} \frac{1}{x+1}$
  Simplify
3. Steps, part 2
1. $\lim_{x \to 0} \frac{1}{x+1} = \frac{1}{0+1} = 1$
  Direct substitution
2. $\lim_{x \to 0} {\left(1+ x\right)}^{\frac{1}{x}} = e^{1} = e$
  Exponentiate the result
4. The answer
$\lim_{x \to 0} {\left(1+ x\right)}^{\frac{1}{x}} = e \approx 2.718281828$
lim x→0 (1 + x)^(1 / x) = e ≈ 2.718281828
5. Check
- Numerical values from the right agree: f(0.01) = 2.7048138, f(0.001) = 2.7169239, f(0.0001) = 2.7181459, f(0.00001) = 2.7182682 ✓
- Numerical values from the left agree: f(-0.01) = 2.731999, f(-0.001) = 2.7196422, f(-0.0001) = 2.7184178, f(-0.00001) = 2.7182954 ✓

Transcript

The problem. Here is the problem: (1 plus x) to the power 1/x). Find the limit of (1 plus x) to the power 1 / x) as x → 0.

Steps, part 1. Indeterminate form 1^∞ — take logarithms. Write the power as e to the power exponent times ln base); since eᵘ is continuous, the limit is e raised to the limit of the exponent. Indeterminate form 0 times ∞ — rewrite as a quotient. Moving one factor to the denominator as its reciprocal turns 0 times ∞ into 0 over 0 or ∞/∞. Indeterminate form 0 over 0 — apply L prime Hôpital prime s rule. Differentiate the numerator and the denominator separately (this is not the quotient rule). Simplify.

Steps, part 2. Direct substitution. The expression is continuous at x equals 0, so the limit equals its value there. Exponentiate the result.

The answer. So the answer is Limit equals e approximately 2.718281828. A limit describes the value a function approaches, not necessarily the value it takes. When substitution gives an indeterminate form, the expression is rewritten (L prime Hôpital prime s rule, a conjugate, or logarithms) until substitution works.

Check. Numerical values from the right agree: f(0.01) equals 2.7048138, f(0.001) equals 2.7169239, f(0.0001) equals 2.7181459, f(0.00001) equals 2.7182682, which checks out. Numerical values from the left agree: f(negative 0.01) equals 2.731999, f(negative 0.001) equals 2.7196422, f(negative 0.0001) equals 2.7184178, f(negative 0.00001) equals 2.7182954, which checks out.

About limits

Evaluate limits by substitution, factoring, standard limits and L'Hôpital's rule.

Find lim x->0 (1 + x)^(1/x): Limits Step-by-Step Lesson