The 1^infinity form arises when lim f(x) = 1 and lim g(x) = infinity. The result is NOT simply 1.

Master Formula: lim f(x)^g(x) = e^(lim g(x) * [f(x) - 1])

Step-by-step method:

1. Confirm 1^infinity form: check base -> 1 and exponent -> infinity
2. Identify f(x) and g(x)
3. Compute f(x) - 1 (this usually simplifies nicely)
4. Multiply by g(x) and evaluate the limit
5. The answer is e^(that limit)

Example: lim(x->0) (cos x)^(1/$x^2$)

• Base: cos(0) = 1, Exponent: 1/0^2 -> infinity. Confirmed 1^infinity.
• f(x) - 1 = cos x - 1
• g(x) * [f(x) - 1] = (1/$x^2$)(cos x - 1) = -$\frac{1 - cos x}{x}$^2
• lim(x->0) -$\frac{1 - cos x}{x}$^2 = -1/2
• Answer: e^(-1/2) = 1/sqrt(e)

JEE frequency: This appears in almost every JEE Main paper. Variations include (1 + 1/n)^n, [$\frac{ax+b}{(cx+d)}$]^(ex+f), and trigonometric bases.