# $1^\infty$-Limit Cases

This is to prove that if $\displaystyle \lim_{x \to \infty} f(x) = 1$ and $\displaystyle \lim_{x \to \infty} g(x) = \infty$, then $\displaystyle \lim_{x \to \infty} f(x)^{g(x)} = e^{\lim_{x \to \infty} g(x)(f(x) - 1)}$.

\begin{aligned} \lim_{x \to \infty} f(x)^{g(x)} & = \lim_{x \to \infty} \left(1+f(x)-1\right)^{g(x)} \\ & = \lim_{x \to \infty} \left(1+\frac 1{\frac 1{f(x)-1}} \right)^{g(x)\left(\frac {f(x)-1}{f(x)-1}\right)} \\ & = \lim_{x \to \infty} \left[\left(1+\frac 1{\frac 1{f(x)-1}} \right)^\frac 1{f(x)-1}\right]^{g(x)(f(x)-1)} \\ & = \lim_{x \to \infty} e^{g(x)(f(x)-1)} \\ & = e^{\lim_{x \to \infty} g(x)(f(x)-1)} \end{aligned}

Note by Chew-Seong Cheong
8 months, 1 week ago

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