11^\infty-Limit Cases

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

limxf(x)g(x)=limx(1+f(x)1)g(x)=limx(1+11f(x)1)g(x)(f(x)1f(x)1)=limx[(1+11f(x)1)1f(x)1]g(x)(f(x)1)=limxeg(x)(f(x)1)=elimxg(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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Hello Sir! How are you? Remember me? I got admitted to an IIT this year!

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I am fine sir, How are you?

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