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Here is the problem: (1 plus x) to the power 1/x).

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Find the limit of (1 plus x) to the power 1 / x) as x → 0.

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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.

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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 ∞/∞.

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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).

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Simplify.

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Direct substitution. The expression is continuous at x equals 0, so the limit equals its value there.

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Exponentiate the result.

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So the answer is Limit equals e approximately 2.718281828.

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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.

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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.

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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.