Does a limit of the form 1^∞ always equal 1?

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The form 1^∞ is indeterminate. It describes a limit in which the base approaches 1 while the exponent grows without bound, and the two effects compete: a base slightly above 1 raised to a huge power can grow large, while the base getting closer to 1 holds it back. The result depends on how fast each happens.

The classic example is the limit of (1 + 1/n)ⁿ as n → ∞, which equals e ≈ 2.71828. The numbers show it: n = 1 gives 2, n = 10 gives about 2.594, and n = 1000 gives about 2.717. The values level off at e, not at 1.

Compare: (1 + 1/n)^(n²) → ∞, while (1 + 1/n²)ⁿ → 1. All three limits have the form 1^∞, with three different answers, which is exactly what "indeterminate" means. Only an exact 1 raised to a power is always 1, and in these limits the base is never exactly 1.

The standard JEE shortcut: if f(x) → 1 and g(x) → ∞, then lim f(x)^g(x) = e^(lim g(x)·(f(x) − 1)). Check it on our example: g·(f − 1) = n × (1/n) = 1, so the limit is e¹ = e. Another example: the limit of (1 + 3x)^(1/x) as x → 0 is e^(lim (1/x)(3x)) = e³. The other indeterminate forms to watch for are 0/0, ∞/∞, 0·∞, ∞ − ∞, 0⁰ and ∞⁰.