Thursday, 27 July 2017

calculus - Indeterminate form $1^infty$ vs. $0^infty$




Why is $1^\infty$ an indeterminate form while $0^\infty = 0$? If $0\cdot0\cdot0\cdots = 0$ shouldn't $1\cdot1\cdot1\cdots = 1$?


Answer



To say that $1^\infty$ is an indeterminate form means that there is more than one object that can be $\lim\limits_{x\,\to\,\text{something}} f(x)^{g(x)}$ where $f(x)\to1$ and $g(x)\to\infty,$ so that the limit depends on which functions $f$ and $g$ are.



Thus
$$
\left.
\begin{align}
& \lim_{x\to\infty} \left(1+\frac 1 x\right) = 1 \quad\text{and} \quad \lim_{x\to\infty} \left( 1 + \frac 1 x \right)^x = e \\[10pt]

& \qquad \text{and} \\[10pt]
& \lim_{x\to\infty} \left( 1 - \frac 1 x\right) = 1 \quad \text{and} \quad \lim_{x\to\infty} \left( 1 - \frac 1 x\right)^x = \frac 1 e.
\end{align} \right\} \longleftarrow \text{two different numbers}
$$


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