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Find lim x->infinity (1 + 2/x)^x: animated lesson

A narrated, step-by-step limits video lesson with captions. Final answer: Limit = e² ≈ 7.389056099.

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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 (0.9 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
$\lim_{x \to \infty} {\left(1+\frac{2}{x}\right)}^{x}$
2. Steps, part 1
1. $\lim_{x \to \infty} {\left(1+\frac{2}{x}\right)}^{x} = e^{\,\lim_{x \to \infty} \left[ x \ln\left(1+\frac{2}{x}\right)\right]}$
  Indeterminate form 1^∞ — take logarithms
2. $\lim_{x \to \infty} x \ln\left(1+\frac{2}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}$
  Indeterminate form 0·∞ — rewrite as a quotient
3. $\lim_{x \to \infty} \frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}} = \lim_{x \to \infty} \frac{-\frac{\frac{2}{{x}^{2}}}{\frac{2}{x}+1}}{-\frac{1}{{x}^{2}}}$
  Indeterminate form 0/0 — apply L'Hôpital's rule
4. $\lim_{x \to \infty} \frac{-\frac{\frac{2}{{x}^{2}}}{\frac{2}{x}+1}}{-\frac{1}{{x}^{2}}} = \lim_{x \to \infty} \frac{2}{\frac{2}{x}+1}$
  Simplify
3. Steps, part 2
1. $\lim_{x \to \infty} \frac{2}{\frac{2}{x}+1} = \lim_{t \to 0} \frac{2}{2\mathrm{t}+1} = 2$
  Substitute x = 1/t (t → 0⁺)
2. $\lim_{x \to \infty} {\left(1+\frac{2}{x}\right)}^{x} = e^{2} = {e}^{2}$
  Exponentiate the result
4. The answer
$\lim_{x \to \infty} {\left(1+\frac{2}{x}\right)}^{x} = {e}^{2} \approx 7.389056099$
lim x→∞ (1 + 2 / x)^x = e² ≈ 7.389056099
5. Check
- Numerical values as x grows agree: f(127.18281) = 7.2749544, f(1271.8281) = 7.3774578, f(12718.281) = 7.3878944, f(127182.81) = 7.3889399 ✓

Transcript

The problem. Here is the problem: (1 plus 2/x) to the power x. Find the limit of (1 plus 2 / x) to the power x as x → ∞.

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. Substitute x equals 1/t (t → 0 to the power plus ). Exponentiate the result.

The answer. So the answer is Limit equals e squared approximately 7.389056099. 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 as x grows agree: f(127.18281) equals 7.2749544, f(1271.8281) equals 7.3774578, f(12718.281) equals 7.3878944, f(127182.81) equals 7.3889399, which checks out.

About limits

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

Find lim x->infinity (1 + 2/x)^x: Limits Step-by-Step Lesson